Difference between revisions of "2003 AMC 12A Problems/Problem 16"

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== Problem ==
  
== Problem ==
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>A point P is chosen at random in the interior of equilateral triangle <math>ABC</math>. What is the probability that <math>\triangle ABP</math> has a greater area than each of <math>\triangle ACP</math> and <math>\triangle BCP</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
A point P is chosen at random from the equilateral triangle ABC. What is the probability that triangle ACP has a greater area than triangles BCP and ABP?
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<math> \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{2}{3} </math>
  
 
== Solution==
 
== Solution==
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<asy>
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draw((0,10)--(8.660254037844385792,-5)--(-8.660254037844385792,-5)--cycle);
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dot((0,0));
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label("$P$",(0,0),N);
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</asy>
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===Solution 1===
  
Solution 1:
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After we pick point <math>P</math>, we realize that <math>ABC</math> is symmetric for this purpose, and so the probability that <math>ACP</math> is the greatest area, or <math>ABP</math> or <math>BCP</math>, are all the same. Since they add to <math>1</math>, the probability that <math>ACP</math> has the greatest area is <math>\boxed{\mathrm{(C)}\ \dfrac{1}{3}}</math>
  
After we pick point P, we realize that ABC is symmetric for this purpose, and so the probability that ACP is the greatest area, or ABP or BCP, are all the same. Since they add to 1, the probability that ACP has the greatest area is 1/3 (C).
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===Solution 2===
  
Solution 2:
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We will use geometric probability. Let us take point <math>P</math>, and draw the perpendiculars to <math>BC</math>, <math>CA</math>, and <math>AB</math>, and call the feet of these perpendiculars <math>D</math>, <math>E</math>, and <math>F</math> respectively. The area of <math>\triangle ACP</math> is simply <math>\frac{1}{2} * AC * PF</math>. Similarly we can find the area of triangles <math>BCP</math> and <math>ABP</math>. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose <math>P, PD + PE + PF</math> = the height of the triangle. Setting the area of triangle <math>ABP</math> greater than <math>ACP</math> and <math>BCP</math>, we want <math>PF</math> to be the largest of <math>PF</math>, <math>PD</math>, and <math>PE</math>. We then realize that <math>PF = PD = PE</math> when <math>P</math> is the incenter of <math>\triangle ABC</math>. Let us call the incenter of the triangle <math>Q</math>. If we want <math>PF</math> to be the largest of the three, by testing points we realize that <math>P</math> must be in the interior of quadrilateral <math>QDCE</math>. So our probability (using geometric probability) is the area of <math>QDCE</math> divided by the area of <math>ABC</math>. We will now show that the three quadrilaterals, <math>QDCE</math>, <math>QEAF</math>, and <math>QFBD</math> are congruent. As the definition of point <math>Q</math> yields, <math>QF</math> = <math>QD</math> = <math>QE</math>. Since <math>ABC</math> is equilateral, <math>Q</math> is also the circumcenter of <math>\triangle ABC</math>, so <math>QA = QB = QC</math>. By the Pythagorean Theorem, <math>BD = DC = CE = EA = AF = FB</math>. Also, angles <math>BDQ, BFQ, CEQ, CDQ, AFQ</math>, and <math>AEQ</math> are all equal to <math>90^\circ</math>. Angles <math>DBF, FAE, ECD</math> are all equal to <math>60</math> degrees, so it is now clear that quadrilaterals <math>QDCE, QEAF, QFBD</math> are all congruent. Summing up these areas gives us the area of <math>\triangle ABC</math>. <math>QDCE</math> contributes to a third of that area so <math>\frac{[QDCE]}{[ABC]}=\boxed{\mathrm{(C)}\ \dfrac{1}{3}}</math>.
  
We will use an approach of geometric probability to solve this problem. Let us take point P, and draw the perpendiculars to AB, BC, and AC, and call the feet of these perpendiculars D, E, and F respectively. The area of triangle ACP is simply 1/2 * AC * PF. Similarly we can find the area of triangles BCP and ABP. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose P, PD + PE + PF = the height of the triangle. Setting the area of triangle ACP greater than ABP and BCP, we want PF to be the largest of PF, PD, and PE. We then realize that PF = PD = PE when P is the orthocenter of ABC. Let us call the orthocenter of the triangle Q. If we want PF to be the largest of the three, by testing points we realize that P must be in the interior of quadrilateral QFCE. So our probability (using geometric probability) is the area of QFCE divided by the area of ABC. We will now show that the three quadrilaterals, QFCE, QEBD, and QDAF are congruent. As the definition of point Q yields, QF = QD = QE. Since ABC is equilateral, Q is also the circumcenter of ABC, so QA = QB = QC. Using the Pythagorean theorem, BD = DA = AF = FC = CE = EB. Also, angles BDQ, BEQ, CEQ, CFQ, AFQ, and ADQ are all equal to 90 degrees by the definition of an altitude. Also, angles DBE, FCE, DAF are all equal to 60 degrees as equilateral triangles are also equiangular. It is now clear that QFCE, QFAD, QEBD are all congruent. Summing up these areas gives us the area of ABC. QFCE contributes to a third of that area, as they are all congruent, so the ratio of the areas of QFCE to ABC is 1/3 (C).
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==See Also==
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{{AMC12 box|year=2003|ab=A|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 14:27, 6 July 2017

Problem

A point P is chosen at random in the interior of equilateral triangle $ABC$. What is the probability that $\triangle ABP$ has a greater area than each of $\triangle ACP$ and $\triangle BCP$?

$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{2}{3}$

Solution

[asy] draw((0,10)--(8.660254037844385792,-5)--(-8.660254037844385792,-5)--cycle); dot((0,0)); label("$P$",(0,0),N); [/asy]

Solution 1

After we pick point $P$, we realize that $ABC$ is symmetric for this purpose, and so the probability that $ACP$ is the greatest area, or $ABP$ or $BCP$, are all the same. Since they add to $1$, the probability that $ACP$ has the greatest area is $\boxed{\mathrm{(C)}\ \dfrac{1}{3}}$

Solution 2

We will use geometric probability. Let us take point $P$, and draw the perpendiculars to $BC$, $CA$, and $AB$, and call the feet of these perpendiculars $D$, $E$, and $F$ respectively. The area of $\triangle ACP$ is simply $\frac{1}{2} * AC * PF$. Similarly we can find the area of triangles $BCP$ and $ABP$. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose $P, PD + PE + PF$ = the height of the triangle. Setting the area of triangle $ABP$ greater than $ACP$ and $BCP$, we want $PF$ to be the largest of $PF$, $PD$, and $PE$. We then realize that $PF = PD = PE$ when $P$ is the incenter of $\triangle ABC$. Let us call the incenter of the triangle $Q$. If we want $PF$ to be the largest of the three, by testing points we realize that $P$ must be in the interior of quadrilateral $QDCE$. So our probability (using geometric probability) is the area of $QDCE$ divided by the area of $ABC$. We will now show that the three quadrilaterals, $QDCE$, $QEAF$, and $QFBD$ are congruent. As the definition of point $Q$ yields, $QF$ = $QD$ = $QE$. Since $ABC$ is equilateral, $Q$ is also the circumcenter of $\triangle ABC$, so $QA = QB = QC$. By the Pythagorean Theorem, $BD = DC = CE = EA = AF = FB$. Also, angles $BDQ, BFQ, CEQ, CDQ, AFQ$, and $AEQ$ are all equal to $90^\circ$. Angles $DBF, FAE, ECD$ are all equal to $60$ degrees, so it is now clear that quadrilaterals $QDCE, QEAF, QFBD$ are all congruent. Summing up these areas gives us the area of $\triangle ABC$. $QDCE$ contributes to a third of that area so $\frac{[QDCE]}{[ABC]}=\boxed{\mathrm{(C)}\ \dfrac{1}{3}}$.

See Also

2003 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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