Question

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.

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 $m+n+p?$

$81$

$89$

$97$

$105$

$113$

Solution:

Let the diagonals meet at $O$ at angle $\theta.$ The feet of the perpendiculars from $A$ and $C$ to $BD$ are symmetric about $O,$ so $PQ=AC\cos\theta=6;$ likewise $RS=BD\cos\theta=8.$

The parallelogram's area is $\tfrac12\cdot AC\cdot BD\sin\theta=15,$ so $AC\cdot BD\sin\theta=30.$ Then $\dfrac{48\sin\theta}{\cos^2\theta}=30,$ giving $\dfrac{\sin\theta}{\cos^2\theta}=\dfrac58.$

Writing $s=\sin\theta,$ $8s=5(1-s^2)$ gives $s=\dfrac{-4+\sqrt{41}}{5},$ so $\cos^2\theta=1-s^2=\dfrac{8(\sqrt{41}-4)}{25}.$

Then $d^2=BD^2=\dfrac{64}{\cos^2\theta}=8(\sqrt{41}+4)=32+8\sqrt{41},$ so $m+n+p=32+8+41=81.$

Thus, the correct answer is A.

2021 AMC 12B Spring Problem 24

Solution:

Problem 24 in Other Years