2026 Hypatia Contest
Solutions
(Grade 11)

Wednesday, April 1, 2026
(in North America and South America)

Thursday, April 2, 2026
(outside of North American and South America)

1. Since $A$, $B$ and $C$ are collinear, segments $AB$ and $AC$ (and $BC$) have the same slope.
  The slope of segment $AB$ is $\dfrac{5-2}{2-1} = 3$. The slope of segment $AC$ is $\dfrac{c-2}{3-1} = \dfrac{c-2}{2}$.
  Therefore, we have that $3 = \dfrac{c-2}{2}$. Solving for $c$ gives us that $c=8$.
2. Since segment $DF$ has slope $\dfrac{7-0}{0-14} = - \dfrac{1}{2}$ and $y$-intercept $7$, the equation of the line
  through points $D$ and $F$ is $y=-\dfrac{1}{2}x + 7$.
  The coordinates $(x,y)$ of $E$ must be positive integers and $E$ must lie on the line $y=-\dfrac{1}{2}x + 7$. When $x \geq 14$ we have that $y \leq 0$, and when $x < 14$ we have that $y>0$. Therefore $x$ and $y$ are both positive exactly when $0<x<14$. Also, $y$ is an integer exactly when $x$ is even.
  Therefore the number of possible points $E$ is the number of integers $0 < x < 14$ such that $x$ is even. There are $6$ such integers which give the following $6$ possibilities for $E$: $(2,6)$, $(4,5)$, $(6,4)$, $(8,3)$, $(10,2)$, and $(12,1)$.
3. The slope of segment $PQ$ is $\dfrac{n-4-12}{6-15}=\dfrac{n-16}{-9}$.
  The slope of segment $PR$ is $\dfrac{n-12}{18-15}=\dfrac{n-12}{3}$.
  Therefore, points $P$, $Q$ and $R$ are collinear exactly when $\dfrac{n-16}{-9}=\dfrac{n-12}{3}$. This is equivalent to $3n-48 = -9n+108$, which is equivalent to $12n=156$.
  Therefore, the value of $n$ that makes $P$, $Q$ and $R$ collinear is $n=\dfrac{156}{12}=13$.
1. Since $0 < A < B < C < 10$, the largest possible value of $A$ is $7$, and when $A=7$, we must have $B=8$ and $C=9$.
  Therefore the largest possible peak number is $78\, 987$.
  Similarly, the smallest possible peak number is $12\, 321$.
  Therefore the positive difference is $78\, 987-12\,321 = 66\, 666$.
2. Since $36\,245 < ABCBA < 45\,932$, we have $A=4$ or $A=3$.
  If $A=4$, then the first inequality is true for all possible values of $B$ and $C$, so we only need to consider $4BCB4 < 45\,932$. We must then have $B=5$, and then $C$ can be any of $6$, $7$, or $8$, so there are $3$ peak numbers when $A=4$. (They are $45\, 654$, $45 \, 754$, $45 \, 854$.)
  If $A=3$, then the second inequality is true for all possible values of $B$ and $C$, so we only need to consider $36\,245 < 3BCB3$. Then $B$ can be $6$, in which case $7 \leq C \leq 9$, or $B = 7$, in which case $8 \leq C \leq 9$, or $B=8$, in which case we have that $C=9$. Altogether, there are $6$ peak numbers when $A=3$. (They are $36\,763$, $36\,863$, $36\,963$, $37\,873$, $37\,973$, $38\,983$.)
  Putting both cases together, there are a total of $3+6 = 9$ peak numbers satisfying the desired inequalities.
3. A number is a multiple of $15$ exactly when it is a multiple of both $5$ and $3$, and a number is a multiple of $5$ exactly when it has units digit $0$ or $5$. Since $0<A<10$ and $ABCBA$ is a multiple of $15$, we must have $A=5$.
  Furthermore, $5BCB5$ is a multiple of $15$ exactly when $5BCB5$ is a multiple of $3$.
  A positive integer is a multiple of $3$ exactly when the sum of its digits is a multiple of $3$. That is, $5BCB5$ is a multiple of $3$ exactly when $10 + 2B + C$ is a multiple of $3$.
  Since $A=5$ and $A<B<C$, the possible values of $B$ are $6$, $7$ and $8$.
  When $B=6$, $10+12+C$ is a multiple of $3$ precisely when $C=8$ (among the possible values $7$, $8$, $9$ of $C$).
  When $B=7$, $10+14+C$ is a multiple of $3$ precisely when $C=9$.
  Finally, when $B=8$, the only possible value of $C$ is 9, and $10+16+9$ is not a multiple of $3$.
  Therefore, the peak numbers that are a multiple of $15$ are $56\,865$ and $57\,975$.
1. Since $DC$ is a diameter of the circle with radius $36$, we have that $DC = 36(2) = 72$. Similarly, $CE= 10(2) = 20$.
  By Fact 1, we have that $DE = CD - CE = 72-20 = 52$.
2. Extend segment $FG$ to $K$, noting that these three points lie on the same line by Fact 1. Also draw in segment $JG$. These constructions are shown in the diagram below.
  
  We have that $FG = FK - GK = 49-12= 37$. By Fact 2, we have that $\angle GJF = 90\degree$. By the Pythagorean Theorem $FG^2 = JG^2 + FJ^2$. That is, $37^2 = 12^2 + FJ^2$. Therefore $FJ^2 = 1225$, and so $FJ=35$ (since $FJ>0$).
3. We will continue to use Facts 1 and 2, but we will no longer cite them each time.
  First draw in line segment $ZL$ (noting it passes through $X$), and then draw in segment $XM$. These constructions are shown below.
  
  We have that $XZ = LZ - LX = 18 - 5 = 13$. Since $XM=5$, by the Pythagorean Theorem we deduce that $MZ=12$. With this result, erase line segment $ZL$ from the above diagram and draw in line segment $XY$, giving us the diagram below.
  
  Note that $18 = ZV = ZY+ YV = ZY+r$ and therefore $ZY = 18-r$. Therefore $MY = MZ + ZY = 12 + (18-r) = 30-r$.
  Note that $XY= XN+NY = 5 + r$. Now apply the Pythagorean Theorem to $\triangle XMY$. We have the following equivalent equations: \[\begin{align*} XM^2 + MY^2 &= XY^2 \\ 5^2 + (30-r)^2 &= (5+r)^2 \\ 25 + 900 - 60r + r^2 &= 25 + 10r + r^2 \\ 70r &= 900 \\ r &= \dfrac{90}{7}\end{align*}\] which is our final answer.
1. From $(\ast)$ we have that $y^2 - z^2 = (m+2)^2 - (m+1)^2$ and $x^2 - y^2 = (m+1)^2 - m^2$. We then have that \[\begin{align*} k &= (y^2 - z^2 ) - (x^2 - y^2) \\ &= ((m+2)^2 - (m+1)^2) - ( (m+1)^2 - m^2) \\ &= (m^2 + 4m + 4 - m^2 - 2m -1) - ( m^2 + 2m + 1 - m^2) \\ &= (2m+3) - (2m + 1) \\ &= 2\end{align*}\] We showed that $y^2-z^2=2m+3$ and $x^2-y^2 = 2m+1$ and since $m$ is a positive integer it follows that $z<y<x$, which we will use in part (b).
2. We begin with a useful fact: the sum of the first $n$ odd positive integers is equal to $n^2$. That is, $1+3+5+\cdots+(2n-1)=n^2$.
  Note: Contestants were not required to justify this, so we save a justification until the end.
  We are given that $y^2 - z^2$ is the sum of $p$ consecutive odd integers, the largest of which is $2y-1$. Therefore, \[\begin{align*} y^2 - z^2 &= \overbrace{(2y-1) + (2y-3) + (2y-5) + \cdots + (2y - (2p-1))}^{\text{p terms}} \\ &= 2yp - (1+3+5 + \cdots + (2p-1)) \\ &= 2yp - p^2\end{align*}\] where we used the useful fact in the final step. Therefore $z^2 = y^2 - 2yp + p^2 = (y-p)^2$, so $z=y-p$ or $-z = y-p$. We will show in fact that $y-z=p$ (and thus $z=y-p$). To see this, note that $y^2 = 1 + 3 + 5 + \cdots + (2y-1)$ expresses $y^2$ as a sum of $y$ odd integers, and $z^2 = 1 + 3 + 5 + \cdots + (2z-1)$ expresses $z^2$ as a sum of $z$ odd integers, and therefore \[\begin{align*} y^2 - z^2 &= 1 + 3 + 5 + \cdots + (2y-1) - (1 + 3 + 5 + \cdots + (2z-1)) \\ &= \text{sum of $y-z$ consecutive odd integers, the largest of which is $2y-1$}\\ & \text{\ \ \ \ (since $y>z$ as shown in part (a))}\end{align*}\] But $y^2 - z^2$ is also the sum of $p$ consecutive odd integers, the largest of which is $2y-1$, and so $y-z=p$. Similar calculations yield $x^2 - y^2 = 2xq - q^2$ and $x-y=q$ or $x=y+q$. We now finish this question with a series of equivalent equations. \[\begin{align*} y^2 - z^2 &= x^2 - y^2 + 2 & \text{(from part (a))}\\ 2yp - p^2 &= 2xq - q^2 + 2\\ 2yp - p^2 &= 2(y+q)q - q^2 + 2\\ 2yp - p^2 &= 2yq +2q^2 - q^2 + 2 \\ 2y (p-q) &= p^2 + q^2 + 2 \\ y &= \dfrac{p^2 + q^2 + 2}{2(p-q)}\end{align*}\] For the final step, note that $p-q \neq 0$, since if $p=q$, the second to last equation would give $0=p^2 + p^2 + 2$. This is equivalent to $p^2 = -1$, which is not possible for an integer $p$.
  We have obtained an expression for $y$ in terms of $p$ and $q$.
3. We just established that $y = \dfrac{p^2+q^2+2}{2(p-q)}$. We begin by finding some values of $p$ and $q$ for which $y$ is a positive integer. If $p-q$ is odd, then one of $p$, $q$ is even and the other is odd, and then $p^2 + q^2 + 2$ is odd. Since the denominator $2(p-q)$ is even, it follows that $y$ is not an integer when $p-q$ is odd. Therefore $p-q$ is even.
  Suppose $p-q=2$. Then $p=q+2$ and \[y = \dfrac{p^2 + q^2+2}{2(p-q)} = \dfrac{(q+2)^2 + q^2+2}{2((q+2)-q)} = \dfrac{q^2 + 4q + 4 + q^2 + 2}{4} = \dfrac{q^2+2q + 3}{2}\] If $q$ is even, then the numerator is odd, and therefore $y$ is not an integer. Thus we must consider odd values of $q$. From (b) we know that $z=y-p = y-(q+2)$ and $x=y+q$. Furthermore, we have that $m^2 + x^2 = (m+1)^2 + y^2$. Solving for $m$ (the $m^2$ terms subtract out) yields that $m= \frac{1}{2}(x^2-y^2 - 1)$. In the table below, for some odd values of $q$, we determine $x$, $y$, $z$, and $m$:
  
  $q$ $x$ $y$ $z$ $m$
  
  $1$ $4$ $3$ $0$ $3$
  
  $3$ $12$ $9$ $4$ $31$
  
  $5$ $24$ $19$ $12$ $107$
  
  When $q=1$, we have that $3^2+4^2 = 4^2 + 3^2 = 5^2 + 0^2$, but we reject $S=25$ since we require $z>0$.
  When $q=3$, we have that $31^2+12^2 = 32^2 + 9^2 = 33^2 + 4^2 = 1105$.
  When $q=5$, we have that $107^2 + 24^2 = 108^2+19^2 =109^2 + 12^2 = 12\, 025$.
  Therefore $1105$ and $12\, 025$ are two Sorensen numbers.
  
  Editorial Comment: It turns out that $1105$ and $12 \,025$ are the smallest two Sorensen numbers and every odd $q \geq 3$ results in a corresponding Sorensen number.
  
  In part (b), we stated and used a useful fact that we will now restate and prove for completeness.
  
  Useful Fact: The sum of the first $n$ odd positive integers is equal to $n^2$.
  Using the more familiar fact that \[1+2+3+\cdots+n = \dfrac{n(n+1)}{2}\] we have that \[\begin{align*} 1+3+5+\cdots+(2n-1) + n &= (1+1)+(3+1)+(5+1)+\cdots+(2n-1+1 )\\ &= 2 + 4 + 6 + \cdots 2n \\ &= 2 [1+2+3+ \cdots + n] \\ &= 2 \dfrac{n(n+1)}{2} \\ &= n^2 + n \end{align*}\] Subtracting $n$ from both sides results in the desired identity, $1+3+5+ \cdots + (2n-1) = n^2$.

\(q\)	\(x\)	\(y\)	\(z\)	\(m\)
\(1\)	\(4\)	\(3\)	\(0\)	\(3\)
\(3\)	\(12\)	\(9\)	\(4\)	\(31\)
\(5\)	\(24\)	\(19\)	\(12\)	\(107\)

2026 Hypatia Contest Solutions (Grade 11)

2026 Hypatia Contest
Solutions
(Grade 11)