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 $[ADP]=\Lambda$ . $[asy] defaultpen(fontsize(14)+0.6); unitsize(12); real x=11.25; pair B1=origin, D1=(x,0), C1=IP(CR(B1,14),CR(D1,10)), A1=OP(CR(B1,10),CR(D1,2*sqrt(65))), P1=extension(A1,C1,B1,D1); pair A=origin, C=(length(C1-A1),0), B=IP(CR(A,10),CR(C,14)), D=OP(CR(A,2*sqrt(65)),CR(C,10)), P=extension(A,C,B,D); draw(A--B--C--D--A); draw(A--C^^B--D,gray+0.4); dot("$A$",A,dir(A-P)); dot("$B$",B,dir(B-P)); dot("$C$",C,dir(C-P)); dot("$D$",D,dir(D-P)); dot("$P$",P,dir(230)); pen p=fontsize(10); label("$10$",A--B,up,p); label("$10$", C--D, 2*right,p); label("$14$", B--C, N,p); label("$2\sqrt{65}$", A--D, SW,p); label("$x$", A--P,down,p); label("$\rho x$", P--C,down,p); label("$\Delta$",(A+B+P)/3, right,p); label("$\Lambda$",(A+D+P)/3, right,p); [/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 reflection of $D$ across $C$ . Let points $E, E',$ and $H$ be the foot of perpendiculars on $AC$ from $D,D'$ , and $B$ respectively. $\begin{align*} &\qquad AB = CD = CD', \quad \textrm{and} \quad BH = DE = D'E' \\ \Rightarrow &\qquad \triangle ABH = \triangle CDE = \triangle CD'E' \\ \Rightarrow &\qquad \angle BAC = \angle ACD' \\ \Rightarrow &\qquad \triangle ABC = \triangle AD'C \\ \Rightarrow &\qquad BC = AD'. \end{align*}$ 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 $[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.$

vladimir.shelomovskii@gmail.com, vvsss

Solution 9 (Mindless Law of Cosines Bash)

Use your favorite method to get that $P$ is the midpoint of one of the two diagonals (suppose it's the midpoint of $\overline{AC}$ ). From here, let $x=AP=PC, y=BP, z=PD, a=\cos\theta$ where $\theta$ is the angle that the diagonals make. Then we have a system of four equations:

\begin{align*} x^2+y^2+2xya=100 \\ z^2+x^2+2xza=100 \\ x^2+y^2-2xya=196 \\ x^2+z^2-2xza=260 \\ \end{align*}

From these equations we get that \begin{align*} xya=-24 \\ xza=-40 \\ x^2+y^2-48=10 \\ x^2+z^2-80=10 \end{align*} From here we can see that $\frac{z}{y}={5}{3}, z^2-y^2=32,$ so $z=5\sqrt{2}, y=3\sqrt{2}.$ Furthermore, this implies $x=\sqrt{130}$ and $xa=-4\sqrt{2},$ which implies $a=\cos\theta=\frac{4}{\sqrt{65}}.$ Then note that the area of the quadrilateral is $\frac{1}{2}\sin\theta (xy+xz+xz+xy)=\sin\theta (\sqrt{130}\cdot 3\sqrt{2}+\sqrt{130} \cdot 5\sqrt{2})=7\cdot (3\cdot 2+5\cdot 2)=7(6+10)=7\cdot 16=\boxed{112}.$

~Dhillonr25

2018 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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