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Difference between revisions of "2023 SSMO Team Round Problems/Problem 11"
 
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dot("$Y$", y, dir(270));
 
dot("$Y$", y, dir(270));
 
</asy>
 
</asy>
 +
 +
~SMO_Team

Latest revision as of 21:34, 9 September 2025

Problem

Let $ABCD$ be a cyclic quadrilateral such that $AC$ is the diameter. Let $P$ be the orthocenter of $ABD$. Define $X = AB\cup CD$, and $Y = AD\cup BC$. If $AB = 8$, $BC = 1$, and $CD = 4$, suppose $\frac{[CBPD]}{[AXY]}=\frac{m}{n}.$ Find $m+n$.

Solution

Since $AC$ is a diameter, $\angle ABC$ and $\angle ADC$ are right angles. Therefore, since $\angle XBY = \angle XDY = 90^\circ$, quadrilateral $BXYD$ is cyclic.

It is well known that $BCDP$ is a parallelogram, which can be proven through angle chasing.

Using the Sine Area Formula, we have \[[CBPD] = 2 \cdot \frac{1}{2} \cdot CB \cdot CD \cdot \sin\angle BCD\] and \[[AXY] = \frac{1}{2} \cdot AX \cdot AY \cdot \sin\angle BAD.\]

Thus, the ratio of the areas is \[\frac{[CBPD]}{[AXY]} = 2 \cdot \frac{CB \cdot CD}{AX \cdot AY},\] since $\sin\angle BCD = \sin\angle BAD$.

Given that the diameter has length $\sqrt{65}$ and $AD = 7$, let $XC = a$. Then, by Power of a Point from $X$, we have \[a(a + 4) = \left(\sqrt{a^2 - 1}\right)\left(\sqrt{a^2 - 1} + 8\right).\]

Expanding both sides: \begin{align*} a^2 + 4a &= a^2 - 1 + 8\sqrt{a^2 - 1} \\ 4a + 1 &= 8\sqrt{a^2 - 1} \\ 16a^2 + 8a + 1 &= 64a^2 - 64. \end{align*}

Solving gives $XC = \frac{5}{4}$. Since $CX \cdot CD = CY \cdot CB$, it follows that $CY = 5$.

Therefore, $BX = \frac{3}{4}$ and $DY = 3$. Then the area ratio is \[2 \cdot \frac{1 \cdot 4}{\frac{35}{4} \cdot 10} = \frac{16}{175},\] so the final answer is $\boxed{191}$.

[asy] import olympiad; import geometry; size(7cm); pair a, b, c, d, p, x, y; a = (0,0); b = (64/sqrt(65),8/sqrt(65)); c = (sqrt(65),0); d = (49/sqrt(65), -28/sqrt(65)); x = intersectionpoint(line(a,b),line(c,d)); y = intersectionpoint(line(a,d),line(b,c)); p = orthocenter(a,b,d); filldraw(circumcircle(a,b,d), cyan+opacity(0.1), blue); filldraw(a--x--y--cycle, red+opacity(0.3), red); filldraw(p--b--c--d--cycle, green+opacity(0.3), green); draw(b--d, dashed+green); draw(x--c--y, blue); dot("$A$", a, dir(180)); dot("$B$", b, dir(45)); dot("$C$", c, dir(0)); dot("$D$", d, dir(270)); dot("$P$", p, dir(180)); dot("$X$", x, dir(50)); dot("$Y$", y, dir(270)); [/asy]

~SMO_Team

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