2015 Cayley Contest
Solutions
(Grade 10)
Tuesday, February 24, 2015
(in North America and South America)
Wednesday, February 25, 2015
(outside of North American and South America)
Tuesday, February 24, 2015
(in North America and South America)
Wednesday, February 25, 2015
(outside of North American and South America)
©2014 University of Waterloo
Answer: (A)
Answer: (D)
Answer: (C)
Answer: (D)
Answer: (B)
Answer: (B)
Answer: (C)
Answer: (D)
Answer: (C)
Therefore, the ratio of the shaded area to the unshaded area is \(6:3\), which equals \(2:1\).
Solution 2
Consider quadrilateral \(YURV\).
\(YURV\) has three right angles: at \(U\) and \(V\) because \(UY\) and \(VY\) are perpendicular to \(QR\) and \(RS\), respectively, and at \(R\) because \(PQRS\) is a square. Since \(YURV\) has three right angles, then it has four right angles and so is a rectangle.
Since \(RV=UR=1\), then \(YURV\) is actually a square and has side length 1, and so has area \(1^2\), or 1.
Similarly, \(XTRW\) is a square of side length 2, and so has area \(2^2\), or 4.
Since square \(PQRS\) is \(3 \times 3\), then its area is \(3^2\), or 9.
The area of the unshaded region is equal to the difference between the areas of square \(XTRW\) and square \(YURV\), or \(4-1=3\).
Since square \(PQRS\) has area 9 and the area of the unshaded region is 3, then the area of the shaded region is \(9-3=6\).
Finally, the ratio of the shaded area to the unshaded area is \(6:3\), which equals \(2:1\).
Answer: (A)
Answer: (E)
Answer: (E)
Answer: (B)
Since the perimeter of \(PQRS\) is 24, then \[\begin{aligned} PQ+RQ+SR+PS & = 24 \\ (22-2x)+(4x-22)+x+x & = 24 \\ 4x & = 24 \\ x & = 6\end{aligned}\] Therefore, \(x=6\).
Answer: (D)
One way to see this is to note that we obtain each successive factorial by multiplying the previous factorial by an integer. (For example, \(6! = 6(5!)\).)
Thus, if one factorial ends in a 0, then all subsequent factorials will also end in a 0.
Since the ones digit of \(5!\) is 0, then the ones digit of each \(n!\) with \(n>5\) will also be 0.
Alternatively, we note that for each positive integer \(n\), the factorial \(n!\) is the product of the positive integers from \(1\) to \(n\). When \(n \geq 5\), the product represented by \(n!\) includes factors of both 2 and 5, and so has a factor of 10, thus has a ones digit of 0.
Therefore, the ones digit of each of \(6!\), \(7!\), \(8!\), \(9!\), and \(10!\) is 0, and so the ones digit of\(6!+7!+8!+9!+10!\) is 0.
Since the ones digit of \(1!+2!+3!+4!+5!\) is 3 and the ones digit of \(6!+7!+8!+9!+10!\) is 0, then the ones digit of \(1!+2!+3!+4!+5!+6!+7!+8!+9!+10!\) is \(3+0\) or \(3\).
(We can verify, using a calculator, that \(1!+2!+3!+4!+5!+6!+7!+8!+9!+10! = 4\,037\,913\).)
Answer: (B)
Answer: (C)
Answer: (B)
Since the diameter of the semicircle is 20, then its radius is half of this, or 10.
Since \(OS\) and \(OR\) are radii, then \(OS=OR=10\).
Consider \(\triangle OPS\) and \(\triangle OQR\).
Since \(PQRS\) is a rectangle, both triangles are right-angled (at \(P\) and \(Q\)).
Also, \(PS=QR\) (equal sides of the rectangle) and \(OS=OR\) (since they are radii of the circle).
Therefore, \(\triangle OPS\) is congruent to \(\triangle OQR\). (Right-angled triangles with equal hypotenuses and one other pair of equal corresponding sides are congruent.)
Since \(\triangle OPS\) and \(\triangle OQR\) are congruent, then \(OP=OQ\).
Since \(PQ=16\), then \(OP = \frac{1}{2}PQ = 8\).
Finally, since \(\triangle OPS\) is right-angled at \(P\), then we can apply the Pythagorean Theorem to conclude that \(PS = \sqrt{OS^2 - OP^2} = \sqrt{10^2-8^2} = \sqrt{100-64}=6\).
Answer: (A)
Answer: (A)
Answer: (E)
Answer: (D)
Answer: (D)
Answer: (E)
Consider square \(F\). It must be covered with either a horizontal tile (covering \(FG\)) or a vertical tile (covering \(FH\)).
If \(F\) is covered with a vertical tile \(FH\), then \(G\) must also be covered with a vertical tile \(GJ\), since \(G\) is covered and its tile cannot overlap \(TU\).
This gives the configuration in Figure 2.
The remaining \(2 \times 2\) square can be covered in the two ways shown in Figure 3 and Figure 4.
This gives 2 tilings of \(PTUS\) so far.
If \(F\) is covered with a horizontal tile \(FG\), then we focus on \(H\) and \(J\).
Either \(H\) and \(J\) are covered by one horizontal tile \(HJ\) (again leaving a \(2 \times 2\) square that can be covered in 2 ways as above (see Figures 5 and 6)) or \(H\) and \(J\) are each covered by vertical tiles \(HK\) and \(JL\), which means that \(MN\) is covered with 1 horizontal tile (see Figure 7).
So if \(F\) is covered with a horizontal tile, there are \(2+1 = 3\) tilings.
In total, there are \(2+3=5\) possible tilings of the \(2 \times 4\) region \(PTUS\).
Consider now region \(TQRU\).
Suppose that the number of tilings of the \(4 \times 4\) region \(TQRU\) is \(t\).
Then for each of the 5 tilings of \(PTUS\), there are \(t\) tilings of \(TQRU\), so there will be \(5t\) tilings of the entire region \(PQRS\).
Divide \(TQRU\) into \(1 \times 1\) squares and label them as shown in Figure 8. We define \(V\) and \(W\) to be the midpoints of \(TQ\) and \(UR\), respectively.
We consider two cases – either the line \(VW\) is overlapped by a tile, or it isn’t.
If \(VW\) is not overlapped by a tile, then each of \(TVWU\) and \(VQRW\) is a \(2\times 4\) region to be tiled, and so can be tiled in 5 ways, as we saw with \(PTUS\).
In this case, the number of tilings of \(TQRU\) is \(5 \times 5 = 25\).
Suppose that \(VW\) is overlapped by at least one tile.
If \(bc\) is covered by a horizontal tile, then \(ae\) and \(dh\) are covered by vertical tiles.
In this case, either \(fg\) is covered by a horizontal tile (Figure 9), or each of \(f\) and \(g\) is covered by a vertical tile (Figure 10).
In the first case (Figure 9), the upper \(4 \times 2\) region needs to be tiled and there are 5 ways to do this, as above.
In the second case (Figure 10), the remaining tiling is forced to be as shown in Figure 11. Can you see why?
Therefore, if \(bc\) is covered by a horizontal tile, there are \(5+1=6\) tilings.
Suppose that \(bc\) is not covered by a horizontal tile, but \(fg\) is covered by a horizontal tile.
Then \(ab\) and \(cd\) are each covered by horizontal tiles and so \(ei\) and \(hl\) are each covered by vertical tiles and so \(mn\) and \(op\) are each covered by horizontal tiles, and so \(jk\) must be covered by a horizontal tile.
In other words, there is only 1 tiling in this case, shown in Figure 12.
Suppose now that \(bc\) and \(fg\) are not covered by a horizontal tile, but \(jk\) is.
In this case, each of the bottom \(2 \times 2\) squares is tiled in a self-contained way. There are 2 ways to tile each, and so \(4\) ways to tile the pair of squares. (These tilings are each self-contained because if either \(ei\) or \(hl\) is covered by a vertical tile, then the remaining three tiles in the corresponding \(2 \times 2\) square cannot be covered with \(1 \times 2\) tiles.)
Furthermore, \(im\) and \(lp\) must be covered by vertical tiles, meaning that \(no\) is tiled with a horizontal tile, as shown in Figure 13, and so there is only one tiling of the upper rectangle.
Thus, there are \(2 \times 2 \times 1 = 4\) tilings of \(TQRU\) in this case, since the rest of the tiling is determined without choice.
Suppose finally that none of \(bc\), \(fg\), or \(jk\) is covered by a horizontal tile, but \(no\) is.
Then \(im\) and \(lp\) are covered with vertical tiles, which means that \(fj\) and \(gk\) are covered by vertical tiles, giving the diagram in Figure 14.
There is no way to complete this tiling without using a horizontal tile \(bc\).
Therefore, there are no tilings in this case.
Finally, we can now say that there are \(t= 25+6+4+1\) tilings of \(TQRU\).
This means that there are \(5 \times 36 = 180\) ways of tiling the entire \(6 \times 4\) region with the given conditions.
Answer: (A)
\(\triangle XAQ\) is right-angled at \(A\) so \(QX^2 = AX^2 + AQ^2\).
But \(AQ\) is the hypotenuse of right-angled \(\triangle ADQ\), so \(AQ^2=AD^2+DQ^2\).
Thus, \(QX^2 = AX^2 + AD^2 + DQ^2\).
Since \(QX=10\), \(AX=h\), \(AD=CQ=n\), and \(DQ=a-m\), then \(10^2 = h^2 + n^2 + (a-m)^2\).
Similarly, using \(PX=12\), we find that \(12^2 = h^2 + n^2 + m^2\) and using \(RX=8\), we find that \(8^2 = h^2 + (a-n)^2 + (a-m)^2\).
Subtracting \(10^2 = h^2 + n^2 + (a-m)^2\) from \(12^2 = h^2 + n^2 + m^2\), we obtain \(144 - 100 = m^2 - (a-m)^2\) or \(44 = m^2 - (a^2-2am+m^2)\) which gives \(44 = 2am - a^2\) or \(m = \dfrac{44+a^2}{2a}\).
Similarly, subtracting \(8^2 = h^2 + (a-n)^2 + (a-m)^2\) from \(10^2 = h^2 + n^2 + (a-m)^2\) gives \(100 - 64 = n^2 - (a-n)^2\) or \(36 = n^2 - (a^2-2an+n^2)\) which gives \(36 = 2an - a^2\) or \(n = \dfrac{36+a^2}{2a}\).
Substituting these expressions for \(m\) and \(n\) into \(12^2 = h^2 + n^2 + m^2\) gives \[\begin{aligned}
h^2 & = 144 - m^2 - n^2 \\
h^2 & = 144 - \left(\dfrac{44+a^2}{2a}\right)^2 - \left(\dfrac{36+a^2}{2a}\right)^2\end{aligned}\] Recall that we want to maximize \(ah\).
Since \(ah>0\), then maximizing \(ah\) is equivalent to maximizing \((ah)^2 = a^2h^2\), which is equivalent to maximizing \(4a^2h^2\).
From above, \[\begin{aligned}
4a^2h^2 & = 4a^2\left(144 - \left(\dfrac{44+a^2}{2a}\right)^2 - \left(\dfrac{36+a^2}{2a}\right)^2\right)\\[1mm]
& = 4a^2\left(144 - \dfrac{(44+a^2)^2}{4a^2}- \dfrac{(36+a^2)^2}{4a^2}\right)\\[1mm]
& = 576a^2 - (44+a^2)^2 - (36+a^2)^2 \\
& = 576a^2 - (1936 + 88a^2 + a^4) - (1296 + 72a^2 + a^4) \\
& = -2a^4 + 416a^2 - 3232 \\
& = -2(a^4 - 208a^2 + 1616) \\
& = -2((a^2-104)^2 + 1616 - 104^2) \qquad\mbox{(completing the square)}\\
& = -2(a^2-104)^2 - 2(1616-104^2) \\
& = -2(a^2-104)^2 + 18400\end{aligned}\] Since \((a^2-104)^2 \geq 0\), then \(4a^2h^2 \leq 18400\) (with equality when \(a = \sqrt{104}\)).
Therefore, \(a^2h^2 \leq 4600\) and so \(ah \leq \sqrt{4600}\).
This means that maximum possible area of \(PQUT\) is \(\sqrt{4600} = 10\sqrt{46} \approx 67.823\).
Of the given answers, this is closest to 67.82.
Answer: (B)