# Difference between revisions of "2021 AMC 12B Problems/Problem 24"

## Problem

Let $ABCD$ be a parallelogram with area $15$. Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.

$[asy] size(350); defaultpen(linewidth(0.8)+fontsize(11)); real theta = aTan(1.25/2); pair A = 2.5*dir(180+theta), B = (3.35,0), C = -A, D = -B, P = foot(A,B,D), Q = -P, R = foot(B,A,C), S = -R; draw(A--B--C--D--A^^B--D^^R--S^^rightanglemark(A,P,D,6)^^rightanglemark(C,Q,D,6)); draw(B--R^^C--Q^^A--P^^D--S,linetype("4 4")); dot("A",A,dir(270)); dot("B",B,E); dot("C",C,N); dot("D",D,W); dot("P",P,SE); dot("Q",Q,NE); dot("R",R,N); dot("S",S,dir(270)); [/asy]$

Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$

$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$

## Solution

Let $X$ denote the intersection point of the diagonals $AC$ and $BD$. Remark that by symmetry $X$ is the midpoint of both $\overline{PQ}$ and $\overline{RS}$, so $XP = XQ = 3$ and $XR = XS = 4$. Now note that since $\angle APB = \angle ARB = 90^\circ$, quadrilateral $ARPB$ is cyclic, and so $$XR\cdot XA = XP\cdot XB,$$which implies $\tfrac{XA}{XB} = \tfrac{XP}{XR} = \tfrac34$.

Thus let $x> 0$ be such that $XA = 3x$ and $XB = 4x$. Then Pythagorean Theorem on $\triangle APX$ yields $AP = \sqrt{AX^2 - XP^2} = 3\sqrt{x^2-1}$, and so$$[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}.$$Solving this for $x^2$ yields $x^2 = \tfrac12 + \tfrac{\sqrt{41}}8$, and so$$(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.$$The requested answer is $32 + 8 + 41 = \boxed{81}$.

## Solution 2(Trig)

Let $X$ denote the intersection point of the diagonals $AC$ and $BD$, and let $\theta = \angle{COB}$. Then, by the given conditions, $XR = 4$, $XQ = 3$, $[XCB] = \frac{15}{4}$. So, $$XC = \frac{3}{\cos \theta}$$ $$XB \cos \theta = 4$$ $$\frac{1}{2} XB XC \sin \theta = \frac{15}{4}$$ Combine the above 3 equations we get $$\frac{\sin \theta }{\cos^2 \theta} = \frac{5}{8}$$. Since we want to find $d^2 = 4XB^2 = \frac{64}{\cos^2 \theta}$, let $x = \frac{1}{\cos^2 \theta}$, then $$\frac{\sin^2 \theta }{\cos^4 \theta} = \frac{1-\cos ^2 \theta}{\cos^4 \theta} = x^2 - x = \frac{25}{64}$$. Solve, we get $x = \frac{4 + \sqrt{41}}{8}$, so $d^2 = 64x = 32 + 8\sqrt{41}$. $\boxed{81}$

~sequoia

~ pi_is_3.14