2024 Canadian Team Mathematics Contest
Solutions
April 2024
April 2024
©2024 University of Waterloo
There are \(7\) days in a week,
so \(7\times 3=21\) days after May
\(3\) is also Thursday. Since \(3+21=24\), May \(24\) is a Thursday.
Thus, May \(25\) is a Friday, May \(26\) is a Saturday, May \(27\) is a Sunday, May \(28\) is a Monday, and May \(29\) is a Tuesday.
Answer: Tuesday
Observe that \(3^2=9 < 11 < 16 =
4^2\), so \(3 < \sqrt{11} <
4\), which implies that \(-4 <
-\sqrt{11} < -3\).
Similarly, \(5^2 = 25 < 29 < 36 =
6^2\), so \(5 < \sqrt{29} <
6\).
From the calculations above, an integer \(n\) is between \(-\sqrt{11}\) and \(\sqrt{29}\) exactly when \(-3\leq n\leq 5\). There are \(9\) such integers.
Answer: \(9\)
Since \(A\), \(E\), and \(C\) lie on a line, \(\angle AEC=180\degree\), and so \(\angle AED+\angle DEC=180\degree\).
Suppose \(\angle AED=7x\degree\). Then
\(\angle DEC=11x\degree\) because \(\angle AED : \angle DEC\) is \(7:11\). Hence, we conclude that \(7x+11x=180\).
Solving this equation for \(x\) gives
\(x=10\), so \(\angle DEC=11\times
10\degree=110\degree\).
The diagonals of a rectangle have equal length and bisect each other, so
\(DE=CE\).
Therefore, \(\triangle CDE\) is
isosceles with \(\angle EDC=\angle
ECD\).
If \(\angle BDC=y\degree\), then since
\(\angle BDC=\angle EDC\), we have
\(\angle EDC=\angle
ECD=y\degree\).
The angles in \(\triangle CDE\) have a
sum of \(180\degree\), so \(2y+110=180\) or \(y=35\).
Answer: \(35\degree\)
If the common ratio is \(r\),
then \(x=80r\), \(y=xr=80r^2\), \(z=yr=80r^3\), and \(3125=zr=80r^4\).
Rearranging \(3125=80r^4\), we get
\(r^4=\dfrac{3125}{80}=\dfrac{625}{16}=\dfrac{25^2}{4^2}\).
Taking square roots, \(r^2=\pm\dfrac{25}{4}\), but \(r\) is a real number, so \(r^2\geq 0\), which means \(r^2=\dfrac{25}{4}\).
Using that \(y=80r^2\) we get \(y=80\times\dfrac{25}{4}=500\).
Although it was not needed to answer the question, there are two
possible values of \(r\) and they are
\(r=-\dfrac{5}{2}\) and \(r=\dfrac{5}{2}\).
Answer: \(500\)
Since \(x\) and \(y\) are both from \(-5\) to \(5\) inclusive, the possible values for
\(x^2\) and \(y^2\) are \(0^2=0\), \(1^2=1\), \(2^2=4\), \(3^2=9\), \(4^2=16\), and \(5^2=25\).
In the table below, the possible sums of \(x^2\) and \(y^2\) are summarized. In particular, the
integer in the row for \(y^2\) and the
column for \(x^2\) is \(x^2+y^2\).
| 0 | 1 | 4 | 9 | 16 | 25 | |
|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 4 | 9 | 16 | 25 |
| 1 | 1 | 2 | 5 | 10 | 17 | 26 |
| 4 | 4 | 5 | 8 | 13 | 20 | 29 |
| 9 | 9 | 10 | 13 | 18 | 25 | 34 |
| 16 | 16 | 17 | 20 | 25 | 32 | 41 |
| 25 | 25 | 26 | 29 | 34 | 41 | 50 |
If a sum in the table above is of the form \(x^2+y^2\) where neither \(x^2\) nor \(y^2\) is zero, then there are four pairs
\((x,y)\) leading to that sum.
For example, the integer \(20\) coming
from \(x^2=4\) and \(y^2=16\) is achieved by \((x,y)\) equal to each of \((2,4)\), \((2,-4)\), \((-2,4)\), and \((-2,-4)\).
If a sum in the table is of the form \(x^2+y^2\) where exactly one of \(x^2\) and \(y^2\) is \(0\), then there are two pairs \((x,y)\) leading to that sum.
For example, the sum of \(16\) coming
from \(x^2=0\) and \(y^2=16\) comes from \((0,4)\) and \((0,-4)\).
There are \(17\) sums in the table
above that are from \(9\) through \(25\).
Of these, \(11\) come from adding two
non-zero squares, and \(6\) come from
adding \(0^2=0\) to a nonzero
square.
From the reasoning above, there are \(2\times
6+4\times 11=56\) pairs \((x,y)\) with the given properties.
Answer: \(56\)
Solution 1
Let \(P\) be on \(AN\) so that \(MP\) is perpendicular to \(AN\), as shown below.
Since \(ABCD\) is a square, \(\angle DAB=90\degree\), which means \(\angle MAP + \angle NAB =
90\degree\).
Since \(\triangle ABN\) is right-angled
at \(B\), we have \(\angle NAB + \angle ANB =
90\degree\).
From \(\angle NAB + \angle ANB = 90\degree =
\angle MAP + \angle NAB\), we can deduce that \(\angle ANB = \angle MAP\).
As well, \(\angle MPA = \angle
ABN=90\degree\), so we conclude that \(\triangle MPA\) and \(\triangle ABN\) are similar.
Since \(ABCD\) is a square, \(BC=AB=2\), and since \(N\) is the midpoint of \(BC\), we have \(BN=1\).
Applying the Pythagorean theorem to \(\triangle ABN\) gives \(AN =
\sqrt{AB^2+BN^2}=\sqrt{2^2+1^2}=\sqrt{5}\).
By the similarity of \(\triangle MPA\)
and \(\triangle ABN\), we have that
\(\dfrac{MP}{AM} =
\dfrac{AB}{AN}=\dfrac{2}{\sqrt{5}}\) or \(MP=\dfrac{2AM}{\sqrt{5}}\). \(M\) is the midpoint of \(AD\), so \(AM=1\), hence \(MP = \dfrac{2}{\sqrt{5}}\).
The lines \(AN\) and \(MC\) are parallel, so the diameter of the
circle is the perpendicular distance between these two lines, which is
the length of \(MP\).
Therefore, the radius of the circle is \(\dfrac{MP}{2}=\dfrac{1}{\sqrt{5}}\) and its
area is \(\pi\left(\dfrac{1}{\sqrt{5}}\right)^2=\dfrac{\pi}{5}\).
Solution 2
We will coordinatize to get \(A(0,2)\), \(B(2,2)\), \(C(2,0)\), and \(D(0,0)\), from which it follows that \(M\) is at \((0,1)\) since it is the midpoint of \(AD\). Similarly, \(N\) is at \((2,1)\).
Line segment \(AN\) has slope \(\dfrac{1-2}{2-0}=-\dfrac{1}{2}\) and line
segment \(MC\) has slope \(\dfrac{0-1}{2-0}=-\dfrac{1}{2}\).
Therefore, \(AN\) and \(MC\) are parallel. By symmetry and because
these lines are parallel, the centre of the circle must be on the line
of slope \(-\dfrac{1}{2}\) that is half
way between \(AN\) and \(MC\).
This line must pass through the midpoint of \(AM\), which is \(\left(0,\dfrac{3}{2}\right)\) so the centre
of the circle lies somewhere on the line with equation \(y=-\dfrac{1}{2}x+\dfrac{3}{2}\).
Therefore, for some \(a\), the centre
of the circle is at \(\left(a,\dfrac{3-a}{2}\right)\).
The circle has equation \(\left(x-a\right)^2+\left(y-\dfrac{3-a}{2}\right)^2=r^2\)
where \(r\) is the radius of the
circle.
The line through \(M(0,1)\) and \(C(2,0)\) has equation \(y=-\dfrac{1}{2}x+1\).
Line segment \(MC\) is tangent to the
circle, so it intersects the circle exactly once.
Substituting \(y=-\dfrac{1}{2}x+1\)
into the equation of the circle gives the following equivalent
equations: \[\begin{align*}
\left(x-a\right)^2+\left(-\dfrac{1}{2}x+1-\dfrac{3-a}{2}\right)^2 &=
r^2 \\
4(x-a)^2 + (-x+2-3+a)^2 &= 4r^2 \\
4(x-a)^2 + (-x+a-1)^2 &= 4r^2 \\
4x^2-8ax+4a^2 + x^2+a^2+1-2ax+2x-2a &= 4r^2 \\
5x^2 + (2-10a)x+5a^2-2a + 1 -4r^2 &= 0\end{align*}\] The
\(x\)-values that satisfy this equation
represent the points of intersection of the circle with \(MC\). There is only one such point, so the
quadratic above must have only one root.
A quadratic with only one root has a discriminant equal to \(0\), so we get the equivalent equations
\[\begin{align*}
(2-10a)^2-4(5)(5a^2-2a+1-4r^2) &= 0 \\
4-40a+100a^2-100a^2+40a-20+80r^2 &= 0 \\
80r^2 &= 16\end{align*}\] Solving \(80r^2=16\) for \(r^2\) gives \(r^2=\dfrac{1}{5}\), so the area of the
circle is \(\pi
r^2=\dfrac{\pi}{5}\).
Answer: \(\dfrac{\pi}{5}\)
The amount of money, in cents, that Rolo has is \(5x + 10y + 25z\).
The amount of money, in cents, that Pat has is \(5y + 10z + 25x\).
The amount of money, in cents, that Sarki has is \(5z + 10x + 25y\).
Therefore, the total amount of money that they have together is \[(5x + 10y + 25z) + (5y + 10z + 25x) + (5z + 10x +
25y) = 40x + 40y + 40z\] They have \(6480\) cents in total, which leads to the
equation \(40(x+y+z)=6480\).
Dividing both sides by \(40\), we get
\(x+y+z=162\).
Since Pat has \(y\) nickels, \(z\) dimes, and \(x\) quarters, she has \(x+y+z=162\) coins in total.
Answer: \(162\)
There are five odd digits: \(1\), \(3\), \(5\), \(7\), and \(9\).
We will count how many integers less than \(1000\) with only odd digits have \(1\) as a units digit.
The integer \(1\) is one of these
integers, and the integers \(11\),
\(31\), \(51\), \(71\), and \(91\) are the two-digit integers with \(1\) as their units digit.
There are \(5\times 5=25\) three-digit
integers with only odd digits and \(1\)
as their units digit. This is because there are \(5\) choices for the hundreds digit and
\(5\) choices for the tens digit, and
these choices are independent.
Thus, \(25+5+1=31\) integers less than
\(1000\) with only odd digits have a
units digit of \(1\).
By nearly identical reasoning, for each possible odd digit, there are
\(31\) integers less than \(1000\) with only odd digits and that units
digit.
When we add all integers less than \(1000\) that have only odd digits, the units
digits contribute \[31(1+3+5+7+9)=31(25)=775\] Doing similar
analysis on the tens digits, we find that for any given odd digit, there
are exactly \(5\) two-digit integers
with only odd digits and that tens digit, and there are \(5\times 5=25\) three-digit integers with
that given tens digit. There are no one-digit integers with an odd tens
digit.
Therefore, for each odd digit, there are \(5+25=30\) integers less than \(1000\) with only odd digits and that given
tens digit.
When we add all integers less than \(1000\) that have only odd digits, the tens
digits contribute \[30(10+30+50+70+90)=7500\] Similarly, there
are \(5\times 5=25\) three-digit
integers with only odd digits for each possible odd hundreds digit. (An
integer with an odd hundreds digit has at least three digits.)
When we add all integers less than \(1000\) that have only odd digits, the
hundreds digits contribute \[25(100+300+500+700+900)=62\,500\]
Therefore, the sum of all integers less than \(1000\) that have only odd digits is \[62\,500+7500+775=70\,775\]
Answer: \(70775\)
Suppose the radius of the sphere is \(r\) and let \(E\) be any point on the circumference of
the circular cross section in the hemisphere containing \(A\).
Then \(\triangle EAO\) has \(\angle EAO=90\degree\) and \(EO=r\). As well, since \(AC=r\), so \(AO=\dfrac{AC}{3}=\dfrac{1}{3}r\).
By the Pythagorean theorem, \(AE^2 + AO^2 =
EO^2\) so \(AE^2=r^2-\left(\dfrac{1}{3}r\right)^2=\dfrac{8}{9}r^2\).
The cone with \(A\) in its base has
base radius \(AE\) and height \(AO\), so its volume is \[\dfrac{1}{3}\pi(AE)^2(AO) =
\dfrac{1}{3}\pi\left(\dfrac{8}{9}r^2\right)\dfrac{r}{3}=\dfrac{8\pi}{81}r^3\]
Let \(F\) be any point on the
circumference of the circular cross section containing \(B\).
Similar to the situation above, we have that \(\triangle FBO\) has \(\angle FBO=90\degree\), \(FO=r\), and \(BO
= \dfrac{2}{3}r\).
By the Pythagorean theorem, we get \(BF^2 =
r^2-\left(\dfrac{2}{3} r\right)^2 = \dfrac{5}{9}r^2\).
The cone with \(B\) in its base has
base radius \(BF\) and height \(BO\), so its volume is \[\dfrac{1}{3}\pi BF^2BO =
\dfrac{1}{3}\pi\left(\dfrac{5}{9}r^2\right)\dfrac{2}{3}r =
\dfrac{10\pi}{81}r^3\] Therefore, the ratio we seek is \(\left(\dfrac{8\pi}{81}r^3\right) :
\left(\dfrac{10\pi}{81}r^3\right)\) which simplifies to \(4:5\).
Answer: \(4:5\)
Label the vertices of the cube by \(A\), \(B\), \(C\), \(D\), \(E\), \(F\), \(G\), and \(H\). The vertices of the cube and the edges connected them can be represented by the following diagram:
Throughout this solution, we will refer to ants by the vertex from
which they originate. For example, Ant \(A\) is the ant that is originally at vertex
\(A\).
Each ant has three choices of which vertex it can walk to. These choices
are summarized in the table below.
| Ant | Choices |
|---|---|
| \(A\) | \(B\), \(D\), \(E\) |
| \(B\) | \(A\), \(C\), \(F\) |
| \(C\) | \(B\), \(D\), \(G\) |
| \(D\) | \(A\), \(C\), \(H\) |
| \(E\) | \(A\), \(F\), \(H\) |
| \(F\) | \(B\), \(E\), \(G\) |
| \(G\) | \(C\), \(F\), \(H\) |
| \(H\) | \(D\), \(E\), \(G\) |
Since each ant has three choices and their choices are independent, there are \(3^8\) possible ways that the \(8\) ants can choose vertices. We will count the number of ways that avoid a collision and get the answer by dividing this total by \(3^8\).
By symmetry, if we assume that Ant \(A\) chooses \(B\), we will count exactly one third of the possibilities. In this case, Ant \(B\) cannot choose \(A\) without colliding with Ant \(A\), so Ant \(B\) must choose \(C\) or \(F\). By symmetry, there are an equal number of choices for each case.
Therefore, we will assume that Ant \(A\) chooses \(B\) and that Ant \(B\) chooses \(C\). This will count exactly one sixth of the choices that avoid a collision.
Ant \(C\) cannot choose \(B\), so it must choose either \(D\) or \(G\). For now, suppose Ant \(C\) chooses \(G\).
The choices for Ant \(F\) are \(B\), \(E\), and \(G\), but \(B\) and \(G\) were chosen by other ants, so Ant \(F\) must choose \(E\).
The choices for Ant \(H\) are \(D\), \(E\), and \(G\), but \(E\) and \(G\) were chosen by other ants, so Ant \(H\) must choose \(D\).
To summarize the choices so far, the table below is the same as the
table above, but with choices crossed out if either the Ant did not
choose that vertex or cannot choose that vertex because another ant has
already chosen it.
| Ant | Choices |
|---|---|
| \(A\) | \(B\), \(\cancel{D}\), \(\cancel{E}\) |
| \(B\) | \(\cancel{A}\), \(C\), \(\cancel{F}\) |
| \(C\) | \(\cancel{B}\), \(\cancel{D}\), \(G\) |
| \(D\) | \(A\), \(\cancel{C}\), \(H\) |
| \(E\) | \(A\), \(F\), \(H\) |
| \(F\) | \(\cancel{B}\), \(E\), \(\cancel{G}\) |
| \(G\) | \(\cancel{C}\), \(F\), \(H\) |
| \(H\) | \(D\), \(\cancel{E}\), \(\cancel{G}\) |
The remaining choices for Ant \(D\)
are \(A\) and \(H\), but Ant \(D\) cannot choose \(H\) without colliding with Ant \(H\) on an edge.
Therefore, Ant \(D\) must choose \(A\), and we can update the table as
follows, eliminating \(A\) as a choice
for Ant \(E\).
| Ant | Choices |
|---|---|
| \(A\) | \(B\), \(\cancel{D}\), \(\cancel{E}\) |
| \(B\) | \(\cancel{A}\), \(C\), \(\cancel{F}\) |
| \(C\) | \(\cancel{B}\), \(\cancel{D}\), \(G\) |
| \(D\) | \(A\), \(\cancel{C}\), \(\cancel{H}\) |
| \(E\) | \(\cancel{A}\), \(F\), \(H\) |
| \(F\) | \(\cancel{B}\), \(E\), \(\cancel{G}\) |
| \(G\) | \(\cancel{C}\), \(F\), \(H\) |
| \(H\) | \(D\), \(\cancel{E}\), \(\cancel{G}\) |
Since Ant \(F\) chose \(E\), Ant \(E\) cannot choose \(F\) since they would collide on an edge.
Therefore, the remaining choices are that Ant \(E\) must choose \(H\) and Ant \(G\) must choose \(F\). One can verify that the choices have
the property that every vertex is chosen by exactly one ant, and no two
ants choose each other’s original vertex, so there is no
collision.
We have now deduced that if Ant \(C\)
chooses \(G\), then there is only one
way for the rest of the ants to choose vertices so that no collision
happens.
From now on, we will assume that Ant \(C\) chooses \(D\). With Ant \(A\) choosing \(B\), and \(B\) choosing \(C\), and Ant \(C\) choosing \(D\), the choices for the ants are
restricted as shown in the table below. Choices have been eliminated if
the vertex was chosen by another ant, or if choosing that vertex would
cause two ants to "swap" vertices and collide on an edge.
| Ant | Choices |
|---|---|
| \(A\) | \(B\), \(\cancel{D}\), \(\cancel{E}\) |
| \(B\) | \(\cancel{A}\), \(C\), \(\cancel{F}\) |
| \(C\) | \(\cancel{B}\), \(D\), \(\cancel{G}\) |
| \(D\) | \(A\), \(\cancel{C}\), \(H\) |
| \(E\) | \(A\), \(F\), \(H\) |
| \(F\) | \(\cancel{B}\), \(E\), \(G\) |
| \(G\) | \(\cancel{C}\), \(F\), \(H\) |
| \(H\) | \(\cancel{D}\), \(E\), \(G\) |
We will now consider the two remaining possibilities for the choice
of Ant \(D\).
If Ant \(D\) chooses \(H\), then the same sort of reasoning that
was used earlier leads to the table below, showing that there is only
one possibility if Ant \(D\) chooses
\(H\).
| Ant | Choices |
|---|---|
| \(A\) | \(B\), \(\cancel{D}\), \(\cancel{E}\) |
| \(B\) | \(\cancel{A}\), \(C\), \(\cancel{F}\) |
| \(C\) | \(\cancel{B}\), \(D\), \(\cancel{G}\) |
| \(D\) | \(\cancel{A}\), \(\cancel{C}\), \(H\) |
| \(E\) | \(A\), \(\cancel{F}\), \(\cancel{H}\) |
| \(F\) | \(\cancel{B}\), \(E\), \(\cancel{G}\) |
| \(G\) | \(\cancel{C}\), \(F\), \(\cancel{H}\) |
| \(H\) | \(\cancel{D}\), \(\cancel{E}\), \(G\) |
Now we assume that Ant \(D\) chooses \(A\) and reduce the possibilities as follows.
| Ant | Choices |
|---|---|
| \(A\) | \(B\), \(\cancel{D}\), \(\cancel{E}\) |
| \(B\) | \(\cancel{A}\), \(C\), \(\cancel{F}\) |
| \(C\) | \(\cancel{B}\), \(D\), \(\cancel{G}\) |
| \(D\) | \(A\), \(\cancel{C}\), \(\cancel{H}\) |
| \(E\) | \(\cancel{A}\), \(F\), \(H\) |
| \(F\) | \(\cancel{B}\), \(E\), \(G\) |
| \(G\) | \(\cancel{C}\), \(F\), \(H\) |
| \(H\) | \(\cancel{D}\), \(E\), \(G\) |
There are two choices for \(E\), and each of them leads to a unique way of the rest of the ants choosing vertices. If Ant \(E\) chooses \(F\), we get Option 1 in the table below, and if Ant \(E\) chooses \(H\), we get Option 2 in the table below.
| Option 1 | Option 2 | ||
|---|---|---|---|
| Ant | Choices | Ant | Choices |
| \(A\) | \(B\), \(\cancel{D}\), \(\cancel{E}\) | \(A\) | \(B\), \(\cancel{D}\), \(\cancel{E}\) |
| \(B\) | \(\cancel{A}\), \(C\), \(\cancel{F}\) | \(B\) | \(\cancel{A}\), \(C\), \(\cancel{F}\) |
| \(C\) | \(\cancel{B}\), \(D\), \(\cancel{G}\) | \(C\) | \(\cancel{B}\), \(D\), \(\cancel{G}\) |
| \(D\) | \(A\), \(\cancel{C}\), \(\cancel{H}\) | \(D\) | \(A\), \(\cancel{C}\), \(\cancel{H}\) |
| \(E\) | \(\cancel{A}\), \(F\), \(\cancel{H}\) | \(E\) | \(\cancel{A}\), \(\cancel{F}\), \(H\) |
| \(F\) | \(\cancel{B}\), \(\cancel{E}\), \(G\) | \(F\) | \(\cancel{B}\), \(E\), \(\cancel{G}\) |
| \(G\) | \(\cancel{C}\), \(\cancel{F}\), \(H\) | \(G\) | \(\cancel{C}\), \(F\), \(\cancel{H}\) |
| \(H\) | \(\cancel{D}\), \(E\), \(\cancel{G}\) | \(H\) | \(\cancel{D}\), \(\cancel{E}\), \(G\) |
To summarize, if we assume that \(A\) chooses \(B\) and \(B\) chooses \(C\), then we will count one sixth of the
possibilities.
Considering cases, if Ant \(C\) chooses
\(G\), then there is exactly one choice
for the rest of the ants that will avoid collisions.
If Ant \(C\) chooses \(D\), then there is one possibility if Ant
\(D\) chooses \(H\), and two possibilities if Ant \(D\) chooses \(A\).
This means \(1+1+2=4\) is one sixth of
the total number of possibilities, so the answer is \[\dfrac{6\times 4}{3^8} =
\dfrac{8}{3^7}=\dfrac{8}{2187}\]
Answer: \(\dfrac{8}{3^7}\) (\(\frac{8}{2187}\))
Since an hour consists of \(3\times
20=60\) minutes, the cyclist will travel \(3\times 8=24\) km in \(1\) hour.
Therefore, the cyclist will travel \(2\times
24=48\) km in two hours.
Answer: \(48~\mathrm{km}\)
Squaring both sides, we get \(x+5 = 5^2=25\), which can be rearranged to get \(x=20\).
Answer: \(20\)
The units digit of \(2A7\times
3\) is the same as the units digit of \(7\times 3=21\), so \(B=1\).
Therefore, \(2A7\times 3 = 711\), so
\(2A7 = \dfrac{711}{3}=237\), so \(A=3\).
The sum of \(A\) and \(B\) is \(A+B=3+1=4\).
Answer: \(4\)
In order to save money, \(k\)
individual rides needs to cost more than \(\$90\).
This means the answer to the question is the smallest integer \(k\) for which \(k\times(\$3.25) > \$90\).
If \(3.25k > 90\), then \(k > \dfrac{90}{3.25} = \dfrac{9000}{325} =
\dfrac{360}{13} = 27 + \dfrac{9}{13}\).
The smallest integer \(k\) such that
\(k > 27 + \dfrac{9}{13}\) is \(k=28\).
Answer: \(28\)
In this solution, we will refer to the squares by their
label.
From the diagram, we can see that E is the only
completely visible square, so it must have been the last to be placed on
the table.
Some of G is covering some of F, so
G must have been placed later than
F.
Some of F is covering some of D, so
F must have been placed later than
D.
By similar reasoning, D must have been placed later
than B, which must have been placed later than both
A and C.
Finally, A is partially covering C, so
A was placed after C.
Putting this all together, the papers were placed in the following
order, given by their labels: C, A,
B, D, F,
G, E.
Answer: CABDFGE
The equation of the first line can be rearranged to get \(6y=2x+42\) or \(y=\dfrac{1}{3}x+7\), so the slope of the
first line is \(\dfrac{1}{3}\).
If \(k=0\), then the other line is
vertical and is not perpendicular to a line with slope \(\dfrac{1}{3}\).
Therefore, the equation of the second line can be rearranged to \(y=-\dfrac{15}{k}x-\dfrac{d}{k}\).
The slopes of perpendicular lines have a product of \(-1\) (unless they are vertical and
horizontal), so \(\left(\dfrac{1}{3}\right)\times\left(\dfrac{-15}{k}\right)
= -1\) or \(-\dfrac{5}{k} =
-1\), so \(k=5\).
The lines intersect on the \(y\)-axis,
which means the two lines intersect when \(x=0\).
Substituting \(k=5\) and \(x=0\) into the two equations, we get \(-6y+42=0\) and \(5y+d=0\).
The equation \(-6y+42=0\) can be solved
to get \(y=7\). Substituting into \(5y+d=0\) gives \(5(7)+d=0\) or \(d=-35\).
Answer: \(-35\)
Suppose \(x=AB\), meaning that
the tens digit of \(x\) is \(A\) and the units digit of \(x\) is \(B\) so that \(x=10A+B\).
Then \(y=BA\) and we have \(18 = y-x = 10B + A - (10A + B) =
9B-9A\).
Dividing both sides by \(9\) gives
\(2=B-A\).
We cannot have \(A=0\) because \(x\) would not be a two-digit integer.
Therefore, the smallest that \(A\) can
be is \(A=1\), which means \(B=A+2=3\).
The integer \(B\) is a digit, so \(B\leq 9\), which implies \(A+2\leq 9\) or \(A\leq 7\).
Therefore, \(A\) can take on the values
\(1\), \(2\), \(3\), \(4\), \(5\), \(6\), and \(7\), for a total of \(7\) possible values of \(A\), and hence, \(7\) possible values of \(x\).
Answer: \(7\)
Let \(P\) be on \(AC\) so that \(BP\) is perpendicular to \(AC\).
Then \(\triangle ADB\) and \(\triangle CDB\) have the same altitude,
\(BP\).
Since they have equal areas and altitudes, they must also have equal
bases, so \(AD=CD\). In other words,
\(D\) is the midpoint of \(AC\).
By a fact about right-triangles that we will prove below, the midpoint
of the hypotenuse is equidistant from the three vertices of the
triangle, so we conclude that \(AD=CD=BD\), but \(AD=\dfrac{AC}{2}=\dfrac{5}{2}\), so \(BD=\dfrac{5}{2}\).
We now prove that if \(\triangle ABC\)
has a right angle at \(B\), then the
midpoint of \(AC\) is equidistant from
\(A\), \(B\), and \(C\).
Suppose \(M\) is the midpoint of \(AC\) and let \(E\) and \(F\) be on \(AB\) and \(BC\), respectively, so that \(AB\) is perpendicular to \(EM\) and \(BC\) is perpendicular to \(FM\).
Since \(BEMF\) has three right
angles, it must have four right angles and be a rectangle, which implies
\(BE = FM\).
Since \(EM\) and \(FC\) are parallel, we get that \(\angle EMA=\angle FCM\).
Since \(EA\) and \(FM\) are parallel, we get that \(\angle EAM = \angle FMC\).
By construction, \(M\) is the midpoint
of \(AC\), so we also have that \(AM = MC\).
Therefore, \(\triangle AEM\) and \(\triangle MFC\) are congruent by
angle-side-angle congruence.
From this congruence, we get that \(EM=FC\), but \(EM=BF\), so we have \(BF=FC\).
Consider \(\triangle MFB\) and \(\triangle MFC\).
These triangles share side \(MF\) and
we have just shown that \(BF=FC\).
We also have that \(MF\) is
perpendicular to \(BC\), so \(\angle BFM=\angle CFM\).
Therefore, \(\triangle BFM\) and \(\triangle CFM\) are congruent by
side-angle-side congruence.
From this congruence, we conclude that \(BM =
CM\), but since \(M\) is the
midpoint of \(AC\), we also have \(CM=AM\), so \(AM=BM=CM\).
Answer: \(\dfrac{5}{2}\)
Let \(m\) be the number rolled
on the \(8\)-sided die and \(n\) be the number rolled on the \(9\)-sided die.
We will count ordered pairs \((m,n)\)
with the property that \(mn\) is a
multiple of \(6\) and divide this total
by \(8\times 9=72\), the number of
possible rolls, to get the probability.
Suppose \(m=1\). Then for \(mn\) to be a multiple of \(6\), we must have that \(n\) is a multiple of \(6\).
The only multiple of \(6\) between
\(1\) and \(9\) inclusive is \(6\), so if \(m=1\), we must have \(n=6\).
Therefore, we get the ordered pair \((1,6)\).
If \(m=5\) or \(m=7\), then we also must have \(n=6\), so we get the pairs \((5,6)\) and \((7,6)\).
If \(m=2\), then \(n\) must be a multiple of \(3\), so \(n=3\), \(n=6\), or \(n=9\). We get \(3\) pairs in this case: \((2,3)\), \((2,6)\), and \((2,9)\).
By similar reasoning, there are three pairs for each of \(m=4\) and \(m=8\).
If \(m=3\), then \(n\) must be even, so \(n=2\), \(n=4\), \(n=6\), or \(n=8\), for a total of \(4\) possibilities.
If \(m=6\), then \(mn\) is always a multiple of \(6\), so there are \(9\) possibilities.
The table below summarizes the work above.
| \(\boldsymbol{m}\) | # of \(\boldsymbol{n}\) for which \(\boldsymbol{mn}\) is a multiple of \(\boldsymbol{6}\) |
|---|---|
| \(1\) | \(1\) |
| \(2\) | \(3\) |
| \(3\) | \(4\) |
| \(4\) | \(3\) |
| \(5\) | \(1\) |
| \(6\) | \(9\) |
| \(7\) | \(1\) |
| \(8\) | \(3\) |
The sum of the numbers in the right column of the table is \(25\), so the probability is \(\dfrac{25}{72}\).
Answer: \(\dfrac{25}{72}\)
The divisors of \(120\) are
\[\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6,
\pm 8, \pm 10, \pm 12, \pm 15, \pm 20, \pm 24, \pm 30, \pm 40, \pm 60,
\pm 120\] of which there are \(32\).
If \(a\) is any of the \(16\) positive divisors of \(120\), then \(b=\dfrac{120}{a}\) is also positive and
\(a+b>0>-23\).
Therefore, for each of the \(16\)
positive divisors \(a\) of \(120\), there is exactly one pair that
satisfies the given conditions.
By similar reasoning, if \(a\) is
negative, then so is \(b\). Therefore,
if either \(a\) or \(b\) is any of \(-24\), \(-30\), \(-40\), \(-60\), or \(-120\), then \(a+b\) will be less than \(-23\).
As well, in any divisor pair of \(120\), one of the two divisors is at least
\(12\) in absolute value.
Therefore, one of \(a\) and \(b\) must be \(-12\), \(-15\), or \(-20\).
If \(a=-12\), then \(b=-10\) and \(a+b=-22 > -23\), so we get the pair
\((a,b)=(-12,-10)\).
If \(a=-15\), then \(b=-8\), so \(a+b=-23\) which is not greater than \(-23\).
If \(a=-20\), then \(b=-6\), so \(a+b
= -26\) which is not greater than \(-23\).
Similarly, if \(b=-12\) then \(a=-10\) and we get the pair \((a,b)=(-10,-12)\), and if \(b=-15\) or \(b=-20\), then \(a+b\) is not greater than \(-23\).
From earlier, there are \(16\) pairs
where \(a\) and \(b\) are both positive. We found two
additional pairs when \(a\) and \(b\) are both negative, and there are no
other pairs, so the answer is \(16+2=18\).
Answer: \(18\)
A parabola with equation \(y=ax^2+bx+c\) with \(a<0\) has its maximum \(y\)-value at \(x=-\dfrac{b}{2a}\).
For the given parabola, the maximum \(y\) value occurs at \(x=-\dfrac{4k}{-4}=k\).
The maximum \(y\)-value is given to be
\(48\), so we must have \[48=-2(k)^2+4k(k)-10k=-2k^2+4k^2-10k=2k^2-10k\]
which can be rearranged to get \(2k^2-10k-48=0\).
Dividing through by \(2\) gives \(k^2-5k-24=0\), which can be factored to get
\((k-8)(k+3)=0\).
Therefore, the possible values of \(k\)
are \(k=8\) and \(k=-3\).
When \(k=8\), the parabola has equation
\(y=-2x^2+32x-80\). This parabola has a
maximum \(y\) value at \(x=8\) and a maximum \(y\) value of \(-2(8)^2+32(8)-80=48\).
When \(k=-3\), the parabola has
equation \(y=-2x^2-12x+30\). This
parabola has a maximum \(y\) value at
\(x=-3\) and a maximum \(y\) value of \(-2(-3)^2-12(-3)+30=48\).
Therefore, the only possible values of \(k\) are \(k=8\) and \(k=-3\), which have a sum of \(5\).
Answer: \(5\)
We want to find \(A+B+C+D\), so
we first set \(x=A+B+C+D\).
The given information leads to the following three equations: \[\begin{align*}
\frac{5}{2}\times\frac{x}{4} &= A \\
\frac{4}{5}\times\frac{x}{4} &= B \\
140 &= C+D\end{align*}\] Adding these equations gives \[\dfrac{5}{8}x + \dfrac{1}{5}x+140 =
A+B+C+D\] Using \(x=A+B+C+D\),
we get \(\dfrac{5}{8}x+\dfrac{1}{5}x+140 =
x\).
Rearranging this equation, we get \(x\left(1-\dfrac{5}{8}-\dfrac{1}{5}\right) =
140\), or \(\dfrac{7}{40}x =
140\).
Multiplying both sides by \(\dfrac{40}{7}\) gives \(x=800\).
Answer: \(800\)
Cubing both sides of the given equation, we get \(\dfrac{3^8+3^n}{3^2+3^n}=3^3\).
Clearing the denominator gives \(3^8+3^n =
3^5+3^33^n\), and rearranging gives \(3^n(3^3-1) = 3^8-3^5\).
Factoring \(3^5\) out of the right side
of the equation above gives \(3^n(3^3-1) =
3^5(3^3-1)\). After dividing by \(3^3-1\), we get \(3^n=3^5\), and so \(n=5\).
Answer: \(5\)
We begin by computing a few terms of the sequence to look for a
pattern.
When \(n=3\), \(n-1=2\) and \(t_2=4\), so \(t_{n-1}\) is even. Therefore, \(t_3=\frac{1}{2}t_2+t_1=2+3=5\).
When \(n=4\), \(t_{n-1}=5\) is odd, so \(t_4 = t_3-t_2=5-4=1\).
When \(n=5\), \(t_{n-1}=1\) is odd, so \(t_5 = t_4-t_3 = 1-5=-4\).
When \(n=6\), \(t_{n-1}=-4\) is even, so \(t_6 = \frac{1}{2}(-4)+1=-1\).
When \(n=7\), \(t_{n-1}=-1\) is odd, so \(t_7 = -1-(-4) = 3\).
When \(n=8\), \(t_{n-1}=3\) is odd, so \(t_8 = 3- (-1) = 4\).
Since each term in the sequence depends only on the previous two terms,
the fact that we have a \(4\) following
a \(3\) again means that the sequence
is periodic. Specifically, the sequence is given by the six terms \(3\), \(4\), \(5\), \(1\), \(-4\), \(-1\) repeating in that order.
Since \(2024 = 6(337)+2\), the first
\(2024\) terms in the sequence are
\(337\) copies of the terms \(3\), \(4\), \(5\), \(1\), \(-4\), and \(-1\), followed by an additional \(3\) and an additional \(4\).
Therefore, the sum of the first \(2024\) terms is \[337(3+4+5+1-4-1) + 3 + 4 =337(8) + 7 =
2703\]
Answer: \(2703\)
For now, consider only the first five letters.
There are two possibilities for the second letter. There are also two
possibilities for the fourth letter since C and E are the only two
letters that can be followed by D.
Therefore, if a sequence of five letters starts with A and ends with D,
then it must be configured in one of the following four ways:
\(\text{A},\text{B},\underline{\ \ \ \ }\,,\text{C},\text{D}\)
\(\text{A},\text{B},\underline{\ \ \ \ }\,,\text{E},\text{D}\)
\(\text{A},\text{C},\underline{\ \ \ \ }\,,\text{C},\text{D}\)
\(\text{A},\text{C},\underline{\ \ \ \ }\,,\text{E},\text{D}\)
The first configuration above is impossible because there is no
letter with the property that it can both follow B and be followed by C.
The third and fourth configurations are impossible by similar
reasoning.
Looking at the second configuration, the only letter that can follow B
and be followed by E is C, and so we conclude that A, B, C, E, D is the
only sequence of five letters starting with A and ending with D.
Therefore, the first five letters must be A, B, C, E, and D in that
order.
Using similar reasoning, since the fifth letter is D and the ninth
letter is A, there are four possible configurations of the fifth through
ninth letters:
\(\text{D},\text{A},\underline{\ \ \ \ }\,,\text{D},\text{A}\)
\(\text{D},\text{A},\underline{\ \ \ \ }\,,\text{E},\text{A}\)
\(\text{D},\text{B},\underline{\ \ \ \ }\,,\text{D},\text{A}\)
\(\text{D},\text{B},\underline{\ \ \ \ }\,,\text{E},\text{A}\)
We will examine these four configurations separately.
For the first configuration, C is the only letter that can follow A and
be followed by D, so one possible sequence is D, A, C, D, A.
For the second configuration, both B and C can follow A and be followed
by E, so we get D, A, B, E, A and D, A, C, E, A as possible
sequences.
For the third configuration, both C and E can follow B and be followed
by D, so we get D, B, C, D, A and D, B, E, D, A as possible
sequences.
For the fourth configuration, C is the only letter that can follow B and
be followed by E , so the only possible sequence is D, B, C, E, A in
this case.
There are six possibilities for the last five letters and only one for
the first five, so there are six possible sequence in total.
Answer: \(6\)
If Ferd removes Tile \(G\), then the board will be configured as shown below, and it can be checked that every available move with this configuration will cause his opponent to lose. Therefore, Ferd will win if he removes Tile \(G\).
We will now show that every other move will give his opponent the ability to guarantee that they win.
Currently, Tiles \(B\), \(D\), \(G\), \(E\), and \(J\) share an edge with an empty square. If
Ferd removes a tile that shares an edge with any of these six tiles,
then he will lose the game.
Therefore, if Ferd removes Tile \(A\),
Tile \(F\), Tile \(H\), or Tile \(K\), he will lose.
Consider the three configurations below. In the first, Tiles \(B\) and \(L\) have been removed, in the second, Tiles
\(E\) and \(L\) have been removed, and in the third,
Tiles \(D\) and \(J\) have been removed.
Notice that in each of the three configurations above, no remaining
tile shares an edge with with at least two empty squares.
As well, it can be verified that removing any tile from any of these
configurations will cause at least one tile to share an edge with at
least two empty squares.
Therefore, if a player is faced with the board in any of the three
configurations above, they will lose the game.
If Ferd removes any of Tiles \(B\),
\(D\), \(E\), \(J\), or \(L\), his opponent can guarantee a
win.
To see why, consider the possibility that Ferd removes Tile \(B\). Then his opponent can remove Tile
\(L\) to leave him in with the first
configuration above.
His opponent will have a similar move if Ferd removes any of Tiles \(D\), \(E\), \(J\), or \(L\), according to one of the three
configurations.
We now have that Ferd will lose (provided his opponent makes the correct
next move) if he removes Tile \(A\),
Tile \(B\), Tile \(D\), Tile \(E\), Tile \(F\), Tile \(H\), Tile \(J\), Tile \(K\), or Tile \(L\). This means he must remove Tile \(G\) to have any chance of winning.
Answer: Tile \(G\)
Using exponent rules, \(4^{\sin^2 x} =
2^{2\sin^2 x}\) and so \(\big(4^{\sin^2
x}\big)\big(2^{\cos^2x}\big) = 2^{2\sin^2x+\cos^2x}\).
As well, \(2\sqrt{2}=2^{\frac{3}{2}}\),
so we have \(2^{2\sin^2x+\cos^2x} =
2^{\frac{3}{2}}\) which implies \(2\sin^2x + \cos^2x = \dfrac{3}{2}\).
Using that \(\sin^2x+\cos^2x=1\), we
have \[\begin{align*}
2\sin^2x + \cos^2x &= \frac{3}{2} \\
\sin^2 x + (\sin^2x+\cos^2x) &= \frac{3}{2} \\
\sin^2x + 1 &= \frac{3}{2}\end{align*}\] from which it
follows that \(\sin^2x=\dfrac{1}{2}\),
and so \(\sin x = \dfrac{1}{\sqrt{2}}\)
or \(\sin x =
-\dfrac{1}{\sqrt{2}}\).
The values of \(x\) with \(0\degree\leq x\leq 360\degree\) for which
\(\sin x=\pm\dfrac{1}{\sqrt{2}}\) are
\(x=45\degree\), \(x=135\degree\), \(x=225\degree\), and \(x=315\degree\).
Every value of \(x\) with \(\sin x=\pm\dfrac{1}{\sqrt{2}}\) is equal to
one of these four values plus a multiple of \(360\degree\).
Since \(315\degree + 5\times 360\degree =
2115\degree\) is larger than \(2024\degree\) but \(315\degree+4\times 360\degree=1755\degree\)
is less than \(2024\degree\), we can
add \(k\times 360\degree\) to the four
angles for \(k=0\), \(k=1\), \(k=2\), \(k=3\), and \(k=4\) and get a value of \(x\) in the range \(0\degree\leq x\leq 2024\degree\).
This gives \(4\times 5=20\) \(x\)-values.
We also get \(x=45\degree+5\times
360\degree=1845\degree\) and \(x=135\degree+5\times
360\degree=1935\degree\) in the desired range, but \(225\degree+5\times 360\degree=2025\degree\)
is too large, so \(x=1935\degree\) is
the largest possible value.
This gives a total of \(20+2=22\)
possible values of \(x\) in the given
range.
Answer: \(22\)
In the diagram below, \(H\) is on \(AD\) so that \(GH\) is perpendicular to \(AD\).
Since \(F\) is the midpoint of \(AB\), we have \(FB = \dfrac{14}{2}=7\).
The height of \(\triangle FGB\) with
base \(FB\) is equal in length to \(AH\).
Since the ratio of the area of \(\triangle
FGB\) to the area of \(ABCD\) is
\(5:28\), we have \[\dfrac{5}{28} = \dfrac{\frac{1}{2}(FB)(AH)}{AB^2}
= \dfrac{\frac{1}{2}(7)(AH)}{14^2} = \dfrac{AH}{56}\] From \(\dfrac{5}{28} = \dfrac{AH}{56}\), we get
\(AH=10\).
Since \(AE=\dfrac{AD}{2}=\dfrac{14}{2}\), we get
\(EH = AH-AE=10-7=3\).
The line segments \(HG\) and \(DC\) are parallel, so \(\angle HGE=\angle DCE\).
As well, \(\angle EHG=\angle
EDC=90\degree\), so we get that \(\triangle EHG\) and \(\triangle EDC\) are similar by angle-angle
similarity.
Thus, we have \(\dfrac{EH}{ED} =
\dfrac{EG}{EC}\), but \(EH=3\)
and \(ED=\dfrac{14}{2}=7\), so \(\dfrac{EG}{EC}=\dfrac{3}{7}\) or \(EC = \dfrac{7}{3}EG\).
If we let \(EG=x\), then we have \(EC=\dfrac{7}{3}x\), and so \(GC =
EC-EG=\dfrac{7}{3}x-x=\dfrac{4}{3}x\).
Therefore, \(EG:GC\) is \(x : \dfrac{4}{3}x\), which can be
simplified to \(3:4\).
Answer: \(3:4\)
Solution 1
Let \(u=x-\dfrac{1}{x}\) and observe that \[\begin{align*} u^3+3u &= \left(x-\frac{1}{x}\right)^3 + 3\left(x-\frac{1}{x}\right) \\ &= x^3-3x+\frac{3}{x}-\frac{1}{x^3} + 3x - \frac{3}{x} \\ &= x^3 - \frac{1}{x^3} \\ &= f(u)\end{align*}\]
So we have that \(f(u)=u^3+3u\),
provided \(u\) is a real number of the
form \(u=x-\dfrac{1}{x}\).
Setting \(1=x-\dfrac{1}{x}\) and
rearranging gives \(x^2-x-1=0\), which
has two real roots since the discriminant of the quadratic is \((-1)^2-4(-1)=5>0\).
Therefore, \(u=1\) is of the given
form, so \(f(1)=1^3+3(1)=4\).
Solution 2
We will first find a value of \(x\)
for which \(1=x-\dfrac{1}{x}\).
Rearranging this equation, we get \(x^2-x-1=0\). By the quadratic formula,
\[x=\dfrac{1\pm\sqrt{(-1)^2-4(-1)}}{2} =
\dfrac{1\pm\sqrt{5}}{2}\]
If \(x=\dfrac{1+\sqrt{5}}{2}\), then \(x-\dfrac{1}{x}=1\) by construction, and we have \[\begin{align*} f(1) &= \left(\frac{1+\sqrt{5}}{2}\right)^3 - \left(\dfrac{2}{1+\sqrt{5}}\right)^3 \\ &= \frac{1+3\sqrt{5}+3\sqrt{5}^2+\sqrt{5}^3}{8} - \dfrac{8}{1+3\sqrt{5}+3\sqrt{5}^2+\sqrt{5}^3} \\ &= \frac{16+8\sqrt{5}}{8} - \dfrac{8}{16+8\sqrt{5}} \\ &= 2+\sqrt{5} - \frac{1}{2+\sqrt{5}} \\ &= 2+\sqrt{5} - \frac{2-\sqrt{5}}{4-5} \\ &= 2+\sqrt{5} + 2 -\sqrt{5} =4\end{align*}\] A similar calculation shows that if \(x=\dfrac{1-\sqrt{5}}{2}\), then \(x^3-\dfrac{1}{x^3}=4\) as well. Therefore \(f(1)=4\).
Answer: \(4\)
We will begin by counting the total number of \(1\)s that are printed when the integers
from \(1\) through \(99\) are printed.
The integers less than \(100\) that
have at least one \(1\) are \[1,10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31,
41, 51, 61, 71, 81, 91\] and a total of \(20\) of the digits written above are \(1\).
When printing the integers from \(100\)
through \(199\), there will be \(100\) 1s coming from the hundreds digit of
each of these integers, plus an additional \(20\) \(1\)s coming from the tens and units digits,
as counted above.
When printing the integers from \(200\)
to \(299\), the digit \(1\) will be printed \(20\) times.
Since the \(100\)s digit is not \(1\) again from \(300\) to \(999\), the digit \(1\) will be printed exactly \(20\) times from \(300\) to \(399\), \(400\) to \(499\), and so on up to \(900\) to \(999\).
Therefore, the number of times the digit \(1\) is printed when the integers from \(1\) through \(999\) are printed is \[9\times 20 + 120 = 300\] When printing the
integers from \(1000\) to \(1099\), the digit \(1\) will be printed once as the thousands
digit for each integer, and an additional \(20\) times for units and tens digits, for a
total of \(120\) times. Adding to the
previous total, in printing the first \(1099\) positive integers, the computer
prints the digit \(1\) a total of \(300+120=420\) times.
When the computer prints the integers from \(1100\) to \(1199\), there will be two \(1\)s for each integer coming from the
thousands and hundreds digits, plus an additional \(20\) \(1\)s, for a total of \(220\) \(1\)s. The computer will have printed \(420+220=640\) \(1\)s when it has printed the first \(1199\) positive integers.
The number of \(1\)s from \(1200\) to \(1299\) is \(120\), and the number of \(1\)s from \(1300\) through \(1399\) is also \(120\), so after the first \(1399\) positive integers have been printed,
\(640+120+120=880\) \(1\)s have been printed.
There are exactly \(120\) \(1\)s from \(1400\) to \(1499\), for a total of \(880+120=1000\) by the time \(1499\) is printed.
Since the integer \(1499\) has one of
its digits equal to \(1\), the \(1000\)th \(1\) is printed when the integer \(1499\) is printed.
Answer: \(1499\)
In this solution, logarithms are base \(10\) unless their base is explicitly
given.
We can rewrite \(x\) as \(x=2024^{\log_{2024}x}\). Using this,
exponent rules, and the change of base formula for logarithms, we get
\[\begin{align*}
4\sqrt{506} &= x^{\log 2024} + 2024^{\log x} \\
&= \left(2024^{\log_{2024}x}\right)^{\log 2024} + 2024^{\log x} \\
&= 2024^{\frac{\log x}{\log 2024}\cdot\log 2024} + 2024^{\log x} \\
&= 2024^{\log x}+2024^{\log x}\end{align*}\] Therefore, we
get \(4\sqrt{506} = 2\cdot 2024^{\log
x}\) or \(2\sqrt{506} = 2024^{\log
x}\).
Noting that \(2\sqrt{506}=2024^{\frac{1}{2}}\), we
conclude that \(\log x = \frac{1}{2}\).
Therefore, \(x=10^{\frac{1}{2}}\) or
\(x=\sqrt{10}\).
Answer: \(\sqrt{10}\)
The unit circle’s position is completely determined by the
location of its centre. Let \(R_1\) be
the region inside the square with the property that if the centre of the
circle is inside \(R_1\), then the
circle will be completely inside the square. Similarly, let \(R_2\) be the region inside the square with
the property that if the centre of the circle is inside \(R_2\), then the circle will intersect
exactly two of the line segments. The probability is the area of \(R_2\) divided by the area of \(R_1\).
In the diagram below, a square of side length \(4\) is centred inside \(ABCD\) so that there is a "strip" of
uniform width \(1\) between the
squares.
Notice that every point inside the smaller square is at least \(1\) unit away from the boundary of \(ABCD\), and every point inside the "strip"
is less than \(1\) unit away from the
boundary of \(ABCD\). Therefore, a
circle of radius \(1\) will be
completely inside \(ABCD\) exactly when
its centre is inside the smaller square.
This means the smaller square is \(R_1\), so the area of \(R_1\) is \(4\times 4=16\).
In the diagram below, a circle of radius \(1\) and a square of side-length \(2\) are centred at \(P\). The vertices of the square are
labelled \(J\), \(K\), \(L\), and \(M\).
Suppose the centre of the circle is outside the square of side-length
\(2\).
If the centre is somewhere above \(JK\), then every point on \(EP\), \(FP\), and \(HP\) is more than one unit away from the
centre of the unit circle, so it is impossible for the unit circle to
intersect any of these line segments.
Therefore, all points above \(JK\) are
outside of \(R_2\).
Similarly, all points below \(LM\), to
the left of \(JM\), and to the right of
\(KL\) are outside of \(R_2\).
These four regions combine to form the entire region outside \(JKLM\), so we conclude that \(R_2\) is contained in \(R_2\).
If the centre of the unit circle is inside the circle centred at \(P\), then \(P\) is inside the unit circle (since both
circles have radius \(1\)). In this
situation, all four of \(EP\), \(FP\), \(GP\), and \(HP\) intersect the unit circle.
Therefore, the region inside the circle centred at \(P\) is not in \(R_2\).
Finally, we suppose the centre of the unit circle is in one of the four
regions inside \(JKLM\) but outside the
circle centred at \(P\).
For example, suppose the centre of the unit circle is in the top-left of
these four regions, with \(J\) as a
vertex.
Then the centre of the circle is within \(1\) unit of \(HP\) and \(GP\), so it must intersect both of them.
However, every point on the interior \(FP\) and \(EP\) is more than \(1\) unit away from the centre of the unit
circle (since all such points are at least as far away as \(P\)).
We have now shown that if the centre of the unit circle is in this small
region, then the unit circle will intersect the interior of \(GP\) and \(HP\), but not \(EP\) or \(FP\).
By similar reasoning, the entire region inside \(JKLM\) and outside the circle centred at
\(P\) is inside \(R_2\). We have also shown that these are
the only possible points in \(R_2\), so
the area of \(R_2\) is the difference
between the area of \(JKLM\) and a
circle of radius \(1\), or \(2\times 2-\pi=4-\pi\).
The probability is \(\dfrac{4-\pi}{4^2}=\dfrac{4-\pi}{16}\).
Answer: \(\dfrac{4-\pi}{16}\)
The number of ways to place zero heads in a row is \(1\). This can only be done by placing six
tails in the row.
The number of ways to place two heads in a row is \(\dbinom{6}{2}=15\).
The number of ways to place four heads in a row is \(\dbinom{6}{4}=15\).
The number of ways to place six heads in a row is \(1\).
Thus, the number of ways to place an even number of heads in a row is
\(1+15+15+1=32=2^5\).
Suppose the first five rows have been arranged to have an even number of
heads in each row. This can be done in \((2^5)^5=2^{25}\) ways.
In order to have an even number of heads in a given column, there is
only one possibility for the coin in the bottom row in that column: if
the first five rows have an even number of heads in that column, then
the sixth must be a tail, and if the first five rows have an odd number
of heads in that column, then the sixth must be a head.
By the previous paragraph, there is exactly one way to assign coins to
the bottom row so that the columns all have an even number of
heads.
Once the coins in the bottom row are arranged, there are an even number
of heads in each column, so the total number of heads in the grid must
be even.
Since the number of heads in the first five rows is even and the total
in all six rows is even, the number of heads in the last row must also
be even.
In other words, if the first five rows are arranged to have an even
number of heads in each row, then there is exactly one way to arrange
the coins in the sixth row so that all conditions are satisfied.
Therefore, the number of ways that the coins can be arranged is \(\left(2^5\right)^5=2^{25}\).
Answer: \(2^{25}\)
The radius of a circle is equal to its circumference divided by
\(2\pi\). Therefore, the radius of the
base is \(\dfrac{3\pi}{2\pi}=\dfrac{3}{2}\), and the
radius of the horizontal cross-section through \(B\) is \(\dfrac{\pi}{2\pi}=\dfrac{1}{2}\).
We will first determine the distance from point \(A\) to \(B\) directly along the surface. This is
equal to the length along the slant height from the base to the
cross-section.
We will draw a perpendicular line from \(B\) to the base of the cone, intersecting
the base at \(D\).
The diagram below is a vertical cross-section of the bottom of the
cone.
The length of \(BD\) is given to be
\(3\sqrt{7}\) and the length of \(AD\) is equal to the difference of the
radii of the base and the cross-section, which is \(\dfrac{3}{2}-\dfrac{1}{2}=1\).
By the Pythagorean theorem, \(AB=\sqrt{(3\sqrt{7})^2+1}=\sqrt{63+1}=8\).
If we let \(E\) be the centre of the
base of the cone, then \(\triangle
AEC\) is similar to \(\triangle
ADB\).
Since \(AE\) is a radius of the base,
we have \(AE=\dfrac{3}{2}\). We saw
earlier that \(AD=1\), so \(\dfrac{AE}{AD}=\dfrac{3}{2}\).
Since \(\triangle AEC\) is similar to
\(\triangle ADB\), \(\dfrac{AC}{AB}=\dfrac{AE}{AD}=\dfrac{3}{2}\),
so \(AC=\dfrac{3}{2}AB=\dfrac{3}{2}(8)=12\).
We will now imagine cutting the top of the cone along line segment \(AC\). This will give a sector of a circle
with the circumference of the base forming the outer arc, and the
circumference of the cross-section through \(B\) forming a smaller sector with the same
angle and centre as the larger sector.
The radius of the larger sector is \(AC=12\), and the length of the arc at the
bottom is equal to the circumference of the base of the cone, which is
\(3\pi\).
The circumference of a circle with radius \(12\) is \(2\times
12\times\pi=24\pi\), and since \(\dfrac{3\pi}{24\pi}=\dfrac{1}{8}\), the
angle of the sector is \(\dfrac{1}{8}\)
of the angle in a full circle, or \(\dfrac{1}{8}\times
360\degree=45\degree\).
The length of \(BC\) is \(12-8=4\), and the shortest path around the
surface of the cone is the line segment connecting \(A\) to \(B\) in the sector.
In \(\triangle ABC\) from the diagram
above, we have \(AC=12\), \(BC=4\), and \(\angle ACB=45\degree\).
Using the Cosine law, we have \[\begin{align*}
p^2 &= AB^2 \\
&= AC^2+BC^2 -2(AC)(AB)\cos\angle ACB\\
&= 12^2 + 4^2 - 2(12)(4)\frac{1}{\sqrt{2}} \\
&= 160-48\sqrt{2}\end{align*}\]
Answer:
\(160-48\sqrt{2}\)
The polynomial factors as \[x^4+2x^3+(3-a^2)x^2+(2-2a^2)x+(1-a^2)=[x^2 +
(1-a)x + (1-a)][x^2 + (1+a)x + (1+a)]\] To find this
factorization, you might first guess that the quartic factors as the
product of two quadratics. The leading coefficient is \(1\) and the constant term factors as \(1-a^2=(1-a)(1+a)\). Looking at other
coefficients, the factorization can be discovered by trial and
error.
The roots of \(x^2+(1-a)x+(1-a)\) are
\[\dfrac{a-1\pm\sqrt{(1-a)^2-4(1-a)}}{2} =
\dfrac{a-1\pm\sqrt{a^2+2a-3}}{2}=\dfrac{a-1\pm\sqrt{(a+1)^2-4}}{2}\]
The roots of \(x^2+(1+a)x+(1+a)\) are
\[\dfrac{-(a+1)\pm\sqrt{(a+1)^2-4(a+1)}}{2}=\dfrac{-(a+1)\pm\sqrt{a^2-2a-3}}{2}=\dfrac{-(a+1)\pm\sqrt{(a-1)^2-4}}{2}\]
A quadratic always has either \(0\),
\(1\), or \(2\) real roots. Since the given quartic is
the product of two quadratics, a root of the quartic must be a root of
one or both of the quadratics.
For the quartic to have exactly \(2\)
real roots, either one quadratic has \(2\) distinct real roots and the other has
no real roots, or each quadratic has \(1\) real root, and those \(2\) roots are different.
Suppose both quadratics have exactly \(1\) real root. This implies both quadratics
have their discriminant equal to \(0\).
Therefore, \((a+1)^2-4=0\) and \((a-1)^2-4=0\). The first equation is
equivalent to \(a+1=\pm 2\) and the
second is equivalent to \(a-1=\pm
2\).
Therefore, the first quadratic has exactly \(1\) real root when \(a=-3\) or \(a=1\), and the second has exactly \(1\) real root when \(a=-1\) or \(a=3\).
Therefore, there is no real number \(a\) for which both quadratics have exactly
\(1\) real root.
We have now shown that the only way for the quartic to have exactly
\(2\) real roots is for one of the
quadratics to have \(2\) real roots and
the other to have no real roots.
Examining discriminants again, we require either \((a+1)^2-4>0\) and \((a-1)^2-4<0\) or \((a+1)^2-4<0\) and \((a-1)^2-4>0\).
Suppose \((a+1)^2-4>0\) and \((a-1)^2-4<0\).
The second of these two inequalities is equivalent to \((a-1)^2<4\), which is equivalent to
\(-2 < a-1 < 2\) or \(-1 < a < 3\).
The first is equivalent to \((a+1)^2>4\), which is equivalent to
\(a+1 < -2\) or \(a+1>2\). These inequalities can be
rearranged to get \(a<-3\) or \(a>1\).
We now conclude that the quartic will have exactly \(2\) real roots if \(-1 < a < 3\) and either \(a<-3\) or \(a>1\).
The inequality \(a<-3\) is
incompatible with the condition \(-1<a<3\), so we must have \(-1<a<3\) and \(a>1\).
Putting these together, we get that \(1<a<3\).
Now suppose \((a+1)^2-4 < 0\) and
\((a-1)^2-4>0\). By similar
reasoning to the previous case, the first of these inequalities implies
\(-3 < a < 1\) and the second
implies either \(a>3\) or \(a<-1\).
This leads to \(-3<a<-1\), but we
only want positive \(a\) values, so the
answer is that the quartic has exactly \(2\) roots when \(1<a<3\).
Answer: \(1 < a < 3\)
(Note: Where possible, the solutions to parts (b) and (c) of each Relay are written as if the value of \(t\) is not initially known, and then \(t\) is substituted at the end.)
The integers in the range that are divisible by \(2\) are \(2,4,6,8,10,12,14,16,18,20\).
The integers in the range that are divisible by \(3\) are \(3,6,9,12,15,18\).
The integers that are in both lists are \(6\), \(12\), and \(18\).
Therefore, the integers that are in exactly one of the two lists are
\[2,3,4,8,9,10,14,15,16,20\] There are
\(10\) integers in the list, so the
answer is \(10\).
Subtracting the second equation from the first, we get \[\begin{align*} (4x+3y) - (-4x+3y) &= 60-(t+2) \\ 8x &= 58-t\end{align*}\] or \(x = \dfrac{58-t}{8}\). Substituting \(t=10\) gives \(x=\dfrac{58-10}{8}=\dfrac{48}{8}=6\).
The surface area of a rectangular prism with dimensions \(a\) by \(b\) by \(c\) is \(2(ab+ac+bc)\).
The surface area of the rectangular prism in the problem is \[2(6r\cdot r + r\cdot t + t\cdot 6r) = 2(6r^2 +
7rt)=12r^2+14rt\] We are given that the surface area is \(18r^2\), and so \(18r^2=12r^2+14rt\) or \(6r^2=14rt\).
Since \(r>0\), we can divide by
\(r\) to get \(6r=14t\) so \(r=\dfrac{14t}{6}=\dfrac{7t}{3}\).
Substituting \(t=6\) gives \(r=\dfrac{7(6)}{3}=14\).
Answer: \((10,6,14)\)
Rearranging the equation, we get \(\dfrac{13}{2}y = -13x + 9\).
Multiplying through by \(\dfrac{2}{13}\) gives \(y=-2x+\dfrac{18}{13}\).
The slope of this line is \(-2\), so a
perpendicular line must have a slope equal to the negative reciprocal of
\(-2\), which is \(\dfrac{-1}{-2}=\dfrac{1}{2}\).
It is given that \(AE=4\) and
\(\dfrac{EB}{AE}=2\), so \(EB=2AE=2\times 4=8\).
Therefore, \(AB=AE+EB=4+8=12\).
Since \(EF\) is parallel to \(BC\), \(\angle
AFE=\angle ACB\) and \(\angle
AEF=\angle ABC\).
Therefore, \(\triangle AEF\) and \(\triangle ABC\) are similar by angle-angle
similarity.
Since these triangles are similar, we get \(\dfrac{AF}{AC} =
\dfrac{AE}{AB}=\dfrac{4}{12}=\dfrac{1}{3}\).
Rearranging \(\dfrac{AF}{AC}=\dfrac{1}{3}\) gives \(3AF=AC\).
We also have that \(AF=AC-FC=AC-10\),
so we can substitute to get \(3(AC-10)=AC\), which can be rearranged to
get \(2AC=30\) or \(AC=15\).
Using that \(\triangle AEF\) is similar
to \(\triangle ABC\), we have \(\dfrac{BC}{EF}=\dfrac{AB}{AE}=\dfrac{12}{4}=3\).
Rearranging \(\dfrac{BC}{EF}=3\) gives
\(BC=3EF=3(4t)=12t\).
We have \(AB=12\), \(AC=15\), and \(BC=12t\), so the perimeter of \(\triangle ABC\) is equal to \(12+15+12t=27+12t\).
Substituting \(t=\dfrac{1}{2}\) gives
\(27+12\times\dfrac{1}{2}=27+6=33\).
The parabola passes through the points with coordinates \((4,5)\) and \((0,5)\).
Since these points have the same \(y\)-coordinate, the axis of symmetry of the
parabola must be at the average of the \(x\)-coordinates, or at \(x=\dfrac{4+0}{2}=2\).
The minimum value of the function occurs on the axis of symmetry, so the
parabola achieves its minimum at \((2,-3)\).
This all implies that the parabola has an equation of the form \(y=a(x-2)^2-3\) for some real number \(a\).
Using that the parabola passes through \((4,5)\), we get \(5=a(4-2)^2-3\) or \(5=4a-3\).
Solving this equation for \(a\) gives
\(a=2\), so the parabola has equation
\(y=2(x-2)^2-3\).
The parabola also passes through \(\left(\dfrac{2t}{11},h\right)\), which
simplifies to \((6,h)\) when \(t=33\) is substituted.
Therefore, \(h=2(6-2)^2-3=2(4)^2-3=29\).
Answer: \((\frac{1}{2},33,29)\)
The sum of the other \(9\) integers is \(83-11=72\). Since there are \(9\) other integers, their average is \(\dfrac{72}{9}=8\).
Since the \(y\)-intercept is
\(-2\), the line passes through the
point \((x,y)=(0,-2)\).
Substituting \((x,y)=(0,-2)\) and \((x,y)=(16,2)\) into the equation \(ax+by=t\) gives \[\begin{align*}
-2b &= t \\
16a+2b &= t\end{align*}\] The first equation is equivalent
to \(b=-\dfrac{t}{2}\), and if we add
the equations we get \(16a=2t\) or
\(a=\dfrac{t}{8}\).
Therefore, \(a-b =
\dfrac{t}{8}-\left(-\dfrac{t}{2}\right) = \dfrac{5t}{8}\).
Substituting \(t=8\) gives \(a-b=5\).
A square with side length \((t+2)\) has area \((t+2)^2\).
The rectangle has area \((4t+8)(t^3+2t^2)\), which can be factored
as \(4(t+2)t^2(t+2)\).
The number of squares needed is the area of the rectangle divided by the
area of the squares, or \[\dfrac{4(t+2)t^2(t+2)}{(t+2)^2}=4t^2\]
Since \(t=5\), the number of squares
needed is \(4(5)^2=100\).
Answer: \(\left(8,5,100\right)\)