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Difference between revisions of "2023 SSMO Team Round Problems/Problem 11"

(Created page with "==Problem== Let <math>ABCD</math> be a cyclic quadrilateral such that <math>AC</math> is the diameter. Let <math>P</math> be the orthocenter of <math>ABD</math>. Define <math>...")
 
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
 +
Since <math>AC</math> is a diameter, <math>\angle ABC</math> and <math>\angle ADC</math> are right angles. Therefore, since <math>\angle XBY = \angle XDY = 90^\circ</math>, quadrilateral <math>BXYD</math> is cyclic.
 +
 +
It is well known that <math>BCDP</math> is a parallelogram, which can be proven through angle chasing.
 +
 +
Using the Sine Area Formula, we have <cmath>[CBPD] = 2 \cdot \frac{1}{2} \cdot CB \cdot CD \cdot \sin\angle BCD</cmath> and <cmath>[AXY] = \frac{1}{2} \cdot AX \cdot AY \cdot \sin\angle BAD.</cmath>
 +
 +
Thus, the ratio of the areas is <cmath>\frac{[CBPD]}{[AXY]} = 2 \cdot \frac{CB \cdot CD}{AX \cdot AY},</cmath> since <math>\sin\angle BCD = \sin\angle BAD</math>.
 +
 +
Given that the diameter has length <math>\sqrt{65}</math> and <math>AD = 7</math>, let <math>XC = a</math>. Then, by Power of a Point from <math>X</math>, we have
 +
<cmath>a(a + 4) = \left(\sqrt{a^2 - 1}\right)\left(\sqrt{a^2 - 1} + 8\right).</cmath>
 +
 +
Expanding both sides:
 +
<cmath>\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*}</cmath>
 +
 +
Solving gives <math>XC = \frac{5}{4}</math>. Since <math>CX \cdot CD = CY \cdot CB</math>, it follows that <math>CY = 5</math>.
 +
 +
Therefore, <math>BX = \frac{3}{4}</math> and <math>DY = 3</math>. Then the area ratio is <cmath>2 \cdot \frac{1 \cdot 4}{\frac{35}{4} \cdot 10} = \frac{16}{175},</cmath> so the final answer is <math>\boxed{191}</math>.
 +
 +
<asy>
 +
        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));
 +
</asy>

Revision as of 21:29, 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 $\boxed{191}$. 

        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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