2023 SSMO Team Round Problems/Problem 11

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 $\boxed{191}$ .

        size(7cm);
        point 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));

        triangle t = triangle(a,b,d); 

        p = orthocenter(t);

        filldraw(circumcircle(t), opacity(0.1)+cyan, blue);
        filldraw(a--x--y--cycle, opacity(0.3)+palered, red);
        filldraw(p--b--c--d--cycle, opacity(0.3)+palegreen, 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));
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2023 SSMO Team Round Problems/Problem 11

Problem

Solution