# Difference between revisions of "2021 AMC 12A Problems/Problem 17"

The following problem is from both the 2021 AMC 10A #17 and 2021 AMC 12A #17, so both problems redirect to this page.

## Problem

Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$, and $\overline{AD}\perp\overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. Given that $OP=11$, the length of $AD$ can be written in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$?

$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$

## Diagram

~MRENTHUSIASM (by Geometry Expressions)

## Solution 1

Angle chasing reveals that $\triangle BPC\sim\triangle BDA$, therefore $$2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43}$$ $$AB=86$$ Additional angle chasing shows that $\triangle ABO \sim\triangle CDO$, therefore $$2=\frac{AB}{CD}=\frac{BP}{PD}=\frac{\frac{BD}{2}+11}{\frac{BD}{2}-11}$$ $$BD=66$$ Since $\triangle ADB$ is right, the Pythagorean theorem implies that $$AD=\sqrt{86^2-66^2}$$ $$AD=4\sqrt{190}$$ $4\sqrt{190}\implies 4 + 190 = \boxed{\textbf{D) } 194}$

~mn28407

## Solution 2 (Similar Triangles, Areas, Pythagorean Theorem)

Since $\triangle BCD$ is isosceles with base $\overline{BD},$ it follows that median $\overline{CP}$ is also an altitude. Let $OD=x$ and $CP=h,$ so $PB=x+11.$

Since $\angle AOD=\angle COP$ by vertical angles, we get $\triangle AOD\sim\triangle COP$ by AA, from which $\frac{AD}{CP}=\frac{OD}{OP},$ or $$AD=CP\cdot\frac{OD}{OP}=h\cdot\frac{x}{11}.$$ Let the brackets denote areas. Notice that $[AOD]=[BOC]$ (By the same base and height, $[ACD]=[BDC].$ Subtracting $[OCD]$ from both sides gives $[AOD]=[BOC].$). Doubling both sides produces \begin{align*} 2[AOD]&=2[BOC] \\ OD\cdot AD&=OB\cdot CP \\ x\left(\frac{hx}{11}\right)&=(x+22)h \\ x^2&=11(x+22). \end{align*} Rearranging and factoring result $(x-22)(x+11)=0,$ from which $x=22.$

Applying the Pythagorean Theorem to right $\triangle CPB,$ we obtain $$h=\sqrt{43^2-33^2}=\sqrt{(43+33)(43-33)}=\sqrt{760}=2\sqrt{190}.$$ Finally, we conclude that $$AD=h\cdot\frac{x}{11}=4\sqrt{190},$$ so the answer is $4+190=\boxed{\textbf{(D) }194}.$

~MRENTHUSIASM

## Solution 3 (Short)

Let $CP = y$ and $CP$ is perpendicular bisector of $DB.$ Let $DO = x,$ so $DP = PB = 11+x.$

(1) $\triangle CPO \sim \triangle ADO,$ so we get $\frac{AD}{x} = \frac{y}{11},$ or $AD = \frac{xy}{11}.$

(2) pythag on $\triangle CDP$ gives $(11+x)^2 + y^2 = 43^2.$

(3) $\triangle BPC \sim \triangle BDA$ with ratio $1:2,$ so $AD = 2y.$ (remember that $P$ is the midpoint of $BD$)

Thus, $xy/11 = 2y,$ or $x = 22.$ And $y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},$ so $AD = 4 \sqrt{190}$ and the answer is $\boxed{194}.$

~ ccx09

## Solution 4 - Extending the line

Observe that $\triangle BPC$ is congruent to $\triangle DPC$; both are similar to $\triangle BDA$. Let's extend $\overline{AD}$ and $\overline{BC}$ past points $D$ and $C$ respectively, such that they intersect at a point $E$. Observe that $\angle BDE$ is $90$ degrees, and that $\angle DBE \cong \angle PBC \cong \angle DBA \implies \angle DBE \cong \angle DBA$. Thus, by ASA, we know that $\triangle ABD \cong \triangle EBD$, thus, $AD = ED$, meaning $D$ is the midpoint of $AE$. Let $M$ be the midpoint of $\overline{DE}$. Note that $\triangle CME$ is congruent to $\triangle BPC$, thus $BC = CE$, meaning $C$ is the midpoint of $\overline{BE}.$

Therefore, $\overline{AC}$ and $\overline{BD}$ are both medians of $\triangle ABE$. This means that $O$ is the centroid of $\triangle ABE$; therefore, because the centroid divides the median in a 2:1 ratio, $\frac{BO}{2} = DO = \frac{BD}{3}$. Recall that $P$ is the midpoint of $BD$; $DP = \frac{BD}{2}$. The question tells us that $OP = 11$; $DP-DO=11$; we can write this in terms of $DB$; $\frac{DB}{2}-\frac{DB}{3} = \frac{DB}{6} = 11 \implies DB = 66$.

We are almost finished. Each side length of $\triangle ABD$ is twice as long as the corresponding side length $\triangle CBP$ or $\triangle CPD$, since those triangles are similar; this means that $AB = 2 \cdot 43 = 86$. Now, by Pythagorean theorem on $\triangle ABD$, $AB^{2} - BD^{2} = AD^{2} \implies 86^{2}-66^{2} = AD^{2} \implies AD = \sqrt{3040} \implies AD = 4 \sqrt{190}$. $4+190 = \boxed{194, \textbf{D}}$

~ ihatemath123

~ pi_is_3.14