2003 AMC 12A Problems/Problem 16

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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 $\triangle BCP$?

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


[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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