2016 AIME II Problems/Problem 14

Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$ .

Solution

The inradius of $\triangle ABC$ is $100\sqrt 3$ and the circumradius is $200 \sqrt 3$ . Now, consider the line perpendicular to plane $ABC$ through the circumcenter of $\triangle ABC$ . Note that $P,Q,O$ must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since $P, Q, O$ are collinear, and $OP=OQ$ , we must have $O$ is the midpoint of $PQ$ . Now, Let $K$ be the circumcenter of $\triangle ABC$ , and $L$ be the foot of the altitude from $A$ to $BC$ . We must have $\tan(\angle KLP+ \angle QLK)= \tan(120^{\circ})$ . Setting $KP=x$ and $KQ=y$ , assuming WLOG $x>y$ , we must have $\tan(120^{\circ})=-\sqrt{3}=\dfrac{\dfrac{x+y}{100 \sqrt{3}}}{\dfrac{30000-xy}{30000}}$ . Therefore, we must have $100(x+y)=xy-30000$ . Also, we must have $\left(\dfrac{x+y}{2}\right)^{2}=\left(\dfrac{x-y}{2}\right)^{2}+120000$ by the Pythagorean theorem, so we have $xy=120000$ , so substituting into the other equation we have $90000=100(x+y)$ , or $x+y=900$ . Since we want $\dfrac{x+y}{2}$ , the desired answer is $\boxed{450}$ .

Solution by Shaddoll

2016 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2016 AIME II Problems/Problem 14

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