2003 AMC 12A Problems/Problem 16

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 $\frac{[QDCE]}{[ABC]}=\boxed{\mathrm{(C)}\ \dfrac{1}{3}}$ .

Solution 3 (Solution 1 but for dumb dumbs)

Due to symmetry, we notice that the probability one triangle is greater than the other is $\frac{1}{2}$ . For two triangles this gives $\frac{1}{2} * \frac{1}{2} = \frac{1}{4}$ . However, the probability of one triangle being greater than the other, depends on knowing whether one triangle is greater or less than the other. This means the probability is less than $\frac{1}{2}$ but greater than $\frac{1}{4}$ giving $(C)$ .

2003 AMC 12A (Problems • Answer Key • Resources)
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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