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 middle point 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 $x^2+y^2-100=-(x^2+y^2-196)$ , 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

2018 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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