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

m (→‎Solution 3 (With yet another way to get the middle point))
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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
  
−
<math>\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</math>.
+
<cmath>\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</cmath>
  
−
What an important finding! Note that the opposite sides <math>\overline{AB}</math> and <math>\overline{CD}</math> have equal length, and note that diagonal <math>\overline{DB}</math> bisects diagonal <math>\overline{AC}</math>. This is very similar to what happens if <math>ABCD</math> were a parallelogram with <math>AB = CD = 10</math>, so let's extend <math>\overline{DB}</math> to point <math>E</math>, such that <math>AECD</math> is a parallelogram. In other words, <math>AE = CD = 10</math> and <math>EC = DA = 2\sqrt{65}</math>. Now, let's examine <math>\triangle ABE</math>. Since <math>AB = AE = 10</math>, the triangle is isosceles, and <math>\angle ABE \cong \angle AEB</math>. Note that in parallelogram <math>AECD</math>, <math>\angle AED</math> and <math>\angle CDE</math> are congruent, so <math>\angle ABE \cong \angle CDE</math> and thus <math>\text{m}\angle ABD = 180^\circ - \text{m}\angle CDB</math>. Define <math>\alpha := \text{m}\angle CDB</math>, so <math>180^\circ - \alpha = \text{m}\angle ABD</math>. We use the Law of Cosines on <math>\triangle DAB</math> and <math>\triangle CDB</math>:
+
What an important finding! Note that the opposite sides <math>\overline{AB}</math> and <math>\overline{CD}</math> have equal length, and note that diagonal <math>\overline{DB}</math> bisects diagonal <math>\overline{AC}</math>. This is very similar to what happens if <math>ABCD</math> were a parallelogram with <math>AB = CD = 10</math>, so let's extend <math>\overline{DB}</math> to point <math>E</math>, such that <math>AECD</math> is a parallelogram. In other words, <cmath>AE = CD = 10</cmath> and <cmath>EC = DA = 2\sqrt{65}</cmath> Now, let's examine <math>\triangle ABE</math>. Since <math>AB = AE = 10</math>, the triangle is isosceles, and <math>\angle ABE \cong \angle AEB</math>. Note that in parallelogram <math>AECD</math>, <math>\angle AED</math> and <math>\angle CDE</math> are congruent, so <math>\angle ABE \cong \angle CDE</math> and thus <cmath>\text{m}\angle ABD = 180^\circ - \text{m}\angle CDB</cmath> Define <math>\alpha := \text{m}\angle CDB</math>, so <math>180^\circ - \alpha = \text{m}\angle ABD</math>.  
  
−
<math>\left(2\sqrt{65}\right)^2 = 10^2 + BD^2 - 20BD\cos\left(180^\circ - \alpha\right) = 100 + BD^2 + 20BD\cos\alpha,</math>
+
We use the Law of Cosines on <math>\triangle DAB</math> and <math>\triangle CDB</math>:
  
−
<math>14^2 = 10^2 + BD^2 - 20BD\cos\alpha.</math>
+
<cmath>\left(2\sqrt{65}\right)^2 = 10^2 + BD^2 - 20BD\cos\left(180^\circ - \alpha\right) = 100 + BD^2 + 20BD\cos\alpha</cmath>
 +
 
 +
<cmath>14^2 = 10^2 + BD^2 - 20BD\cos\alpha</cmath>
  
 
Subtracting the second equation from the first yields
 
Subtracting the second equation from the first yields
  
−
<math>260 - 196 = 40BD\cos\alpha \rightarrow BD\cos\alpha = \frac{8}{5}.</math>
+
<cmath>260 - 196 = 40BD\cos\alpha \rightarrow BD\cos\alpha = \frac{8}{5}</cmath>
  
−
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 <cmath>\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56</cmath> and hence <cmath>\left[ABCD\right] = 56 + 56 = \boxed{112}</cmath> -kgator
  
  

Revision as of 09:01, 5 September 2021

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 midpoint 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, \[(\frac{1}{2}AP\cdot BP+\frac{1}{2}CP\cdot DP)\sin{APB}=(\frac{1}{2}AP\cdot DP+\frac{1}{2}CP\cdot BP)\sin{APD}\] 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

$x^2+y^2-100=2 \cdot x \cdot y \cdot \cos{APB}=-(2 \cdot x \cdot y \cdot \cos{(\pi-CPB)})=-(x^2+y^2-196)$, 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

See Also

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

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