# 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 $ABCD$.

## Diagram

Let $AP=x$ and let $PC=\rho x$. Let $[ABP]=\Delta$ and let $[ADP]=\Lambda$. $[asy] size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A, B, C, D, P; real theta = 10; A = origin; D = dir(theta)*(2*sqrt(65), 0); B = 10*dir(theta + 147.5); C = IP(CR(D,10), CR(B,14)); P = extension(A,C,B,D); draw(A--B--C--D--cycle, s); draw(A--C^^B--D); dot("A", A, SW); dot("B", B, NW); dot("C", C, NE); dot("D", D, SE); dot("P", P, 2*dir(115)); label("10", A--B, SW); label("10", C--D, 2*N); label("14", B--C, N); label("2\sqrt{65}", A--D, S); label("x", A--P, SE); label("\rho x", P--C, SE); label("\Delta", B--P/2, 3*dir(-25)); label("\Lambda", D--P/2, 7*dir(180)); [/asy]$

## Solution 1

Let $AP=x$ and let $PC=\rho x$. Let $[ABP]=\Delta$ and let $[ADP]=\Lambda$. We easily get $[PBC]=\rho \Delta$ and $[PCD]=\rho\Lambda$.

We are given that $[ABP] +[PCD] = [PBC]+[ADP]$, which we can now write as $$\Delta + \rho\Lambda = \rho\Delta + \Lambda \qquad \Longrightarrow \qquad \Delta -\Lambda = \rho (\Delta -\Lambda).$$ Either $\Delta = \Lambda$ or $\rho=1$. The former would imply that $ABCD$ is a parallelogram, which it isn't; therefore we conclude $\rho=1$ and $P$ is the midpoint of $AC$. Let $\angle BAD = \theta$ and $\angle BCD = \phi$. Then $[ABCD]=2\cdot [BCD]=140\sin\phi$. On one hand, since $[ABD]=[BCD]$, we have \begin{align}\sqrt{65}\sin\theta = 7\sin\phi \quad \implies \quad 16+49\cos^2\phi = 65\cos^2\theta\end{align}whereas, on the other hand, using cosine formula to get the length of $BD$, we get $$10^2+4\cdot 65 - 40\sqrt{65}\cos\theta = 10^2+14^2-280\cos\phi$$\begin{align}\tag{2}\implies \qquad 65\cos^2\theta = \left(7\cos\phi+ \frac{8}{5}\right)^2\end{align}Eliminating $\cos\theta$ in the above two equations and solving for $\cos\phi$ we get$$\cos\phi = \frac{3}{5}\qquad \implies \qquad \sin\phi = \frac{4}{5}$$which finally yields $[ABCD]=2\cdot [BCD] = 140\sin\phi = 112$.

## Solution 2

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 -- $h_1$ and $h_2$ can actually be equal. In this case, $AP \neq CP$, but $BP$ must be equal to $DP$. We get the same result. -Mathdummy.

## Solution 3 (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 4 (With yet another way to get the middle point)

Denote $\angle APB$ by $\alpha$. Then $\sin(\angle APB)=\sin \alpha = \sin(\angle APD)$. Using the formula for the area of a triangle, we get $$\frac{1}{2} (AP\cdot BP+ CP\cdot DP)\sin\alpha=\frac{1}{2}(AP\cdot DP+ CP\cdot BP)\sin\alpha ,$$ 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 $\triangle APB$ and $\triangle BPC$, it is clear that $$x^2+y^2-100=2 xy \cdot \cos{APB}=-(2 xy \cdot \cos{(\pi-CPB)})=-(x^2+y^2-196)$$ or \begin{align}x^2+y^2=148\end{align}. Likewise, using the cosine rule for triangles $APD$ and $CPD$, \begin{align}\tag{2}x^2+z^2=180\end{align}. It follows that \begin{align}\tag{3}z^2-y^2=32\end{align}. 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 $$\frac{48^2}{y^2}=\frac{80^2}{z^2} \qquad \Longrightarrow \qquad 5y=3z.$$ 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 5

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 $[DAP]=[DCP]$, and $[BAP]=[BCP] \implies [ABD] = [CBD]$, therefore, \begin{align} \nonumber \frac 12 \cdot AB\cdot AD\sin A &= \frac 12 \cdot BC\cdot CD\sin C \\ \Longrightarrow \hspace{1in} 7\sin C &= \sqrt{65}\sin A \end{align} By Law of Cosines, \begin{align} \nonumber 10^2+14^2-2\cdot 10\cdot 14\cos C &= 10^2+4\cdot 65-2\cdot 10\cdot 2\sqrt{65}\cos A \\ \Longrightarrow \hspace{1in} -\frac 85 - 7\cos C &= \sqrt{65}\cos A \tag{2} \end{align} Square $(1)$ and $(2)$, and add them, to get $$\left(\frac 85\right)^2 + 2\cdot \frac 85 \cdot 7\cos C + 7^2 = 65$$ Solve, $\cos C = 3/5 \implies \sin C = 4/5$, $$[ABCD] = 2[BCD] = BC\cdot CD\cdot \sin C = 14\cdot 10\cdot \frac 45 = \boxed{112}$$ -Mathdummy

## Solution 6

Either $PA=PC$ or $PD=PB$. Let $PD=PB=s$. Applying Stewart's Theorem on $\triangle ABD$ and $\triangle BCD$, dividing by $2s$ and rearranging, $$\tag{1}CP^2+s^2=148$$ $$\tag{2}AP^2+s^2=180$$ Applying Stewart on $\triangle CAB$ and $\triangle CAD$, $$\tag{3} 5CP^2=3AP^2$$ Substituting equations 1 and 2 into 3 and rearranging, $s=BP=PD\sqrt{130}, CP=3\sqrt{2}, PA=5\sqrt{2}$ . By Law of Cosines on $\triangle APB$, $\cos(\angle APB)=\frac{4\sqrt{65}}{65}$ so $\sin(\angle APB)=\sin(\angle BPC)=\sin(\angle CPD)=\sin(\angle DPA)=\frac{7\sqrt{65}}{65}$. Using $[\triangle ABC]=\frac{ab\sin(\angle C)}{2}$ to find unknown areas, $[ABCD]=[\triangle APB]+[\triangle BPC]+[\triangle CPD]+[\triangle DPA]=\boxed{112}$.

-Solution by Gart

## Solution 7

Now we prove P is the midpoint of $BD$. Denote the height from $B$ to $AC$ as $h_1$, height from $D$ to $AC$ as $h_2$.According to the problem, $AP* h_1 +CP* h_2 =CP* h_1 +AP* h_2$ implies $h_1 (AP-CP)= h_2 (AP-CP), h_1 = h_2$. Then according to basic congruent triangles we get $BP=DP$ Firstly, denote that $CP=a,BP=b,CP=c,AP=d$. Applying Stewart theorem, getting that $100c+196b=(b+c)(bc+a^2); 100b+260c=(b+c)(bc+c^2); 100c+196b=100b+260c, 3b=5c$, denote $b=5x,c=3x$ Applying Stewart Theorem, getting $260a+100a=2a(a^2+25x^2); 196a+100a=2a(9x^2+a^2)$ solve for a, getting $a=\sqrt{130},AP=5\sqrt{2}; CP=3\sqrt{2}$ Now everything is clear, we can find $cos\angle{BPA}=\frac{4}{\sqrt{65}}$ using LOC, $sin\angle{BPA}=\frac{7\sqrt{65}}{65}$, the whole area is $\sqrt{130}*8\sqrt{2}*\frac{7\sqrt{65}}{65}=\boxed{112}$

~bluesoul

## Solution 8 (Simple Geometry)

$BP = PD$ as in another solutions.

Let $D'$ be the point symmetrical $D$ with respect to $C.$ Let points $E, E',$ and $H$ be the foot of perpendiculars from $D,D',$ and $B$ to $AC,$ respectively. $$AB = CD = CD', BH = DE = D'E' \implies$$ $$\triangle ABH = \triangle CDE = \triangle CD'E' \implies \angle BAC = \angle ACD'$$ $$\implies \triangle ABC = \triangle AD'C \implies BC = AD'.$$ The area of quadrilateral $ABCD$ is equal to the area of triangle $ADD'$ with sides $AD' = 14, AD = 2\sqrt{65}, DD' = 2 \cdot 10 = 20$. The semiperimeter is $s = 17 + \sqrt{65},$ the area $$[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.$$