# 2018 AIME II Problems/Problem 12

## Problem

Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$, $BC = 14$, and $AD = 2\sqrt{65}$. Assume that the diagonals of $ABCD$ intersect at point $P$, and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$. Find the area of quadrilateral $ABCD$.

## Solution 1

For reference, $2\sqrt{65} \approx 16$, so $\overline{AD}$ is the longest of the four sides of $ABCD$. Let $h_1$ be the length of the altitude from $B$ to $\overline{AC}$, and let $h_2$ be the length of the altitude from $D$ to $\overline{AC}$. Then, the triangle area equation becomes

$\frac{h_1}{2}AP + \frac{h_2}{2}CP = \frac{h_1}{2}CP + \frac{h_2}{2}AP \rightarrow \left(h_1 - h_2\right)AP = \left(h_1 - h_2\right)CP \rightarrow AP = CP$.

What an important finding! Note that the opposite sides $\overline{AB}$ and $\overline{CD}$ have equal length, and note that diagonal $\overline{DB}$ bisects diagonal $\overline{AC}$. This is very similar to what happens if $ABCD$ were a parallelogram with $AB = CD = 10$, so let's extend $\overline{DB}$ to point $E$, such that $AECD$ is a parallelogram. In other words, $AE = CD = 10$ and $EC = DA = 2\sqrt{65}$. Now, let's examine $\triangle ABE$. Since $AB = AE = 10$, the triangle is isosceles, and $\angle ABE \cong \angle AEB$. Note that in parallelogram $AECD$, $\angle AED$ and $\angle CDE$ are congruent, so $\angle ABE \cong \angle CDE$ and thus $\text{m}\angle ABD = 180^\circ - \text{m}\angle CDB$. Define $\alpha := \text{m}\angle CDB$, so $180^\circ - \alpha = \text{m}\angle ABD$. We use the Law of Cosines on $\triangle DAB$ and $\triangle CDB$:

$\left(2\sqrt{65}\right)^2 = 10^2 + BD^2 - 20BD\cos\left(180^\circ - \alpha\right) = 100 + BD^2 + 20BD\cos\alpha,$

$14^2 = 10^2 + BD^2 - 20BD\cos\alpha.$

Subtracting the second equation from the first yields

$260 - 196 = 40BD\cos\alpha \rightarrow BD\cos\alpha = \frac{8}{5}.$

This means that dropping an altitude from $B$ to some foot $Q$ on $\overline{CD}$ gives $DQ = \frac{8}{5}$ and therefore $CQ = \frac{42}{5}$. Seeing that $CQ = \frac{3}{5}\cdot BC$, we conclude that $\triangle QCB$ is a 3-4-5 right triangle, so $BQ = \frac{56}{5}$. Then, the area of $\triangle BCD$ is $\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56$. Since $AP = CP$, points $A$ and $C$ are equidistant from $\overline{BD}$, so $\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56$ and hence $\left[ABCD\right] = 56 + 56 = \boxed{112}$. -kgator

Just to be complete -- $h1$ and $h2$ can actually be equal. In this case, $AP \neq CP$, but $BP$ must be equal to $DP$. We get the same result. -Mathdummy.

## Solution 2 (Another way to get the middle point)

So, let the area of $4$ triangles $\triangle {ABP}=S_{1}$, $\triangle {BCP}=S_{2}$, $\triangle {CDP}=S_{3}$, $\triangle {DAP}=S_{4}$. Suppose $S_{1}>S_{3}$ and $S_{2}>S_{4}$, then it is easy to show that $$S_{1}\cdot S_{3}=S_{2}\cdot S_{4}.$$ Also, because $$S_{1}+S_{3}=S_{2}+S_{4},$$ we will have $$(S_{1}+S_{3})^2=(S_{2}+S_{4})^2.$$ So $$(S_{1}+S_{3})^2=S_{1}^2+S_{3}^2+2\cdot S_{1}\cdot S_{3}=(S_{2}+S_{4})^2=S_{2}^2+S_{4}^2+2\cdot S_{2}\cdot S_{4}.$$ So $$S_{1}^2+S_{3}^2=S_{2}^2+S_{4}^2.$$ So $$S_{1}^2+S_{3}^2-2\cdot S_{1}\cdot S_{3}=S_{2}^2+S_{4}^2-2\cdot S_{2}\cdot S_{4}.$$ So $$(S_{1}-S_{3})^2=(S_{2}-S_{4})^2.$$ As a result, $$S_{1}-S_{3}=S_{2}-S_{4}.$$ Then, we have $$S_{1}+S_{4}=S_{2}+S_{3}.$$ Combine the condition $$S_{1}+S_{3}=S_{2}+S_{4},$$ we can find out that $$S_{3}=S_{4},$$ so $P$ is the midpoint of $\overline {AC}$

~Solution by $BladeRunnerAUG$ (Frank FYC)

## Solution 3 (With yet another way to get the middle point)

Using the formula for the area of a triangle, $$(\frac{1}{2}AP.BP+\frac{1}{2}CP.BP)\sin{APB}=(\frac{1}{2}AP.BP+\frac{1}{2}CP.BP)\sin{APD}$$ But $\sin{APB}=\sin{APD}$, so $$(AP-CP)(BP-DP)=0$$ Hence $AP=CP$ (note that $BP=DP$ makes no difference here). Now, assume that $AP=CP=x$,$BP=y$, and $DP=z$. Using the cosine rule for triangles $APB$ and $BPC$, it is clear that

$x^2+y^2-100=2 \cdot x \cdot y \cdot \cos{APB}=-(2 \cdot x \cdot y \cdot \cos{(\pi-CPB)})=-(x^2+y^2-196)$, or $$x^2+y^2=148...(1)$$ Likewise, using the cosine rule for triangles $APD$ and $CPD$, $$x^2+z^2=180...(2)$$. It follows that $$z^2-y^2=32...(3)$$. Now, denote angle $APB$ by $\alpha$. Since $\sin\alpha=\sqrt{1-\cos^2\alpha}$, $$\sqrt{1-\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\sqrt{1-\frac{(x^2+z^2-260)^2}{4x^2z^2}}$$ which simplifies to $$\frac{48^2}{y^2}=\frac{80^2}{z^2}$$, giving $$5y=3z$$. Plugging this back to equations (1), (2), and (3), it can be solved that $x=\sqrt{130},y=3\sqrt{2},z=5\sqrt{2}$. Then, the area of the quadrilateral is $$x(y+z)\sin\alpha=\sqrt{130}\cdot8\sqrt{2}\cdot\frac{14}{\sqrt{260}}=\boxed{112}$$ --Solution by MicGu

## Solution 4

As in all other solutions, we can first find that either $AP=CP$ or $BP=DP$, but it's an AIME problem, we can take $AP=CP$, and assume the other choice will lead to the same result (which is true).

From $AP=CP$, we have $[DAP]=[DCP]$, $[BAP]=[BCP]$ => $[ABD] = [CBD]$, therefore, $$1/2AB*AD\sin A = 1/2BC*CD\sin C => 7\sin C = \sqrt{65}\sin A ... (1)$$ By Law of Cosine, $$10^2+14^2-2*10*14\cos C=10^2+4*65-2*10*2\sqrt{65}\cos A$$ $$(-8/5)-7\cos C = \sqrt{65}\cos A ...(2)$$ Square (1) and (2), add them, we get $$(8/5)^2 +2(8/5)7\cos C + 7^2 = 65$$ Solve, $\cos C = 3/5$ => $\sin C = 4/5$, $$[ABCD] = 2[BCD] = BC*CD*\sin C = 14*10*(4/5) = \boxed{112}$$ -Mathdummy