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2018 AIME II Problems/Problem 12

Revision as of 00:06, 11 February 2019 by Micgu (talk | contribs) (→‎Solution 3 (With yet another way to get the middle point))

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

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