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Difference between revisions of "2018 AIME II Problems/Problem 12"

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Let <math>ABCD</math> be a convex quadrilateral with <math>AB = CD = 10</math>, <math>BC = 14</math>, and <math>AD = 2\sqrt{65}</math>. Assume that the diagonals of <math>ABCD</math> intersect at point <math>P</math>, and that the sum of the areas of triangles <math>APB</math> and <math>CPD</math> equals the sum of the areas of triangles <math>BPC</math> and <math>APD</math>. Find the area of quadrilateral <math>ABCD</math>.
 
Let <math>ABCD</math> be a convex quadrilateral with <math>AB = CD = 10</math>, <math>BC = 14</math>, and <math>AD = 2\sqrt{65}</math>. Assume that the diagonals of <math>ABCD</math> intersect at point <math>P</math>, and that the sum of the areas of triangles <math>APB</math> and <math>CPD</math> equals the sum of the areas of triangles <math>BPC</math> and <math>APD</math>. Find the area of quadrilateral <math>ABCD</math>.
  
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==Solution==
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==Solution 1==
  
 
For reference, <math>2\sqrt{65} \approx 16</math>, so <math>\overline{AD}</math> is the longest of the four sides of <math>ABCD</math>. Let <math>h_1</math> be the length of the altitude from <math>B</math> to <math>\overline{AC}</math>, and let <math>h_2</math> be the length of the altitude from <math>D</math> to <math>\overline{AC}</math>. Then, the triangle area equation becomes
 
For reference, <math>2\sqrt{65} \approx 16</math>, so <math>\overline{AD}</math> is the longest of the four sides of <math>ABCD</math>. Let <math>h_1</math> be the length of the altitude from <math>B</math> to <math>\overline{AC}</math>, and let <math>h_2</math> be the length of the altitude from <math>D</math> to <math>\overline{AC}</math>. Then, the triangle area equation becomes
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This means that dropping an altitude from <math>B</math> to some foot <math>Q</math> on <math>\overline{CD}</math> gives <math>DQ = \frac{8}{5}</math> and therefore <math>CQ = \frac{42}{5}</math>. Seeing that <math>CQ = \frac{3}{5}\cdot BC</math>, we conclude that <math>\triangle QCB</math> is a 3-4-5 right triangle, so <math>BQ = \frac{56}{5}</math>. Then, the area of <math>\triangle BCD</math> is <math>\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56</math>. Since <math>AP = CP</math>, points <math>A</math> and <math>C</math> are equidistant from <math>\overline{BD}</math>, so <math>\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56</math> and hence <math>\left[ABCD\right] = 56 + 56 = \boxed{112}</math>. -kgator
 
This means that dropping an altitude from <math>B</math> to some foot <math>Q</math> on <math>\overline{CD}</math> gives <math>DQ = \frac{8}{5}</math> and therefore <math>CQ = \frac{42}{5}</math>. Seeing that <math>CQ = \frac{3}{5}\cdot BC</math>, we conclude that <math>\triangle QCB</math> is a 3-4-5 right triangle, so <math>BQ = \frac{56}{5}</math>. Then, the area of <math>\triangle BCD</math> is <math>\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56</math>. Since <math>AP = CP</math>, points <math>A</math> and <math>C</math> are equidistant from <math>\overline{BD}</math>, so <math>\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56</math> and hence <math>\left[ABCD\right] = 56 + 56 = \boxed{112}</math>. -kgator
  
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==Solution 2 (Another way to get the middle point)==
  
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So, let the area of <math>4</math> triangles <math>\triangle {ABP}=S_{1}</math>,  <math>\triangle {BCP}=S_{2}</math>,  <math>\triangle {CDP}=S_{3}</math>,  <math>\triangle {DAP}=S_{4}</math>. Suppose <math>S_{1}>S_{3}</math> and <math>S_{2}>S_{4}</math>, then it is easy to show that <cmath>S_{1}\cdot S_{3}=S_{2}\cdot S_{4}</cmath>. Also, because <cmath>S_{1}+S_{3}=S_{2}+S_{4}</cmath>, we will have <cmath>(S_{1}+S_{3})^2=(S_{2}+S_{4})^2</cmath>. So <cmath>(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}</cmath>. So <cmath>S_{1}^2+S_{3}^2=S_{2}^2+S_{4}^2</cmath>. So <cmath>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}</cmath>. So <cmath>(S_{1}-S_{3})^2=(S_{2}-S_{4})^2</cmath>. As a result, <cmath>S_{1}-S_{3}=S_{2}-S_{4}</cmath>. Then, we have <cmath>S_{1}+S_{4}=S_{2}+S_{3}</cmath>. Combine the condition <cmath>S_{1}+S_{3}=S_{2}+S_{4}</cmath>, we can find out that <cmath>S_{3}=S_{4}</cmath>. So <math>P</math> is the middle point of <math>\overline {AC}</math>
  
 
{{AIME box|year=2018|n=II|num-b=11|num-a=13}}
 
{{AIME box|year=2018|n=II|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:00, 6 October 2018

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

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 $\overline {AC}$

2018 AIME II (Problems • Answer Key • Resources)
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Problem 13
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