Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 2 (Another way to get the middle point))
Line 24: Line 24:
  
 
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>
 
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>
 +
 +
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)
  
 
{{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 21:01, 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 $\overline {AC}$

~Solution by $BladeRunnerAUG$ (Frank FYC)

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
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_12&oldid=98041"