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Difference between revisions of "2016 AIME II Problems/Problem 14"

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

Revision as of 16:20, 22 March 2018

Problem

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 1

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

Short Simple Solution

Draw a good diagram. Draw $CH$ as an altitude of the triangle. Scale everything down by a factor of $100\sqrt{3}$, so that $AB=2\sqrt{3}$. Finally, call the center of the triangle U. Draw a cross-section of the triangle via line $CH$, which of course includes $P, Q$. From there, we can call $OU=h$. There are two crucial equations we can thus generate. WLOG set $PU<QU$, then we call $PU=d-h, QU=d+h$. First equation: using the Pythagorean Theorem on $\triangle UOB$, $h^2+2^2=d^2$. Next, using the tangent addition formula on angles $\angle PHU, \angle UHQ$ we see that after simplifying $-d^2+h^2=-4, 2d=3\sqrt{3}$ in the numerator, so $d=\frac{3\sqrt{3}}{2}$. Multiply back the scalar and you get $\boxed{450}$. Not that hard, was it?

Solution 3

To make numbers more feasible, we'll scale everything down by a factor of $100$ so that $\overline{AB}=\overline{BC}=\overline{AC}=6$. We should also note that $P$ and $Q$ must lie on the line that is perpendicular to the plane of $ABC$ and also passes through the circumcenter of $ABC$ (due to $P$ and $Q$ being equidistant from $A$, $B$, $C$), let $D$ be the altitude from $C$ to $AB$. We can draw a vertical cross-section of the figure then: [asy]pair C, D, I, P, Q, O; D=(0,0); C=(5.196152,0); P=(1.732051,7.37228); I=(1.732051,0); Q=(1.732051,-1.62772); O=(1.732051,2.87228); draw(C--Q--D--P--cycle); draw(C--D, dashed); draw(P--Q, dotted); draw(O--C, dotted); label("$C$", C, E); label("$D$", D, W); label("$I$", I, NW); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, SW); dot(O); dot(I);[/asy] We let $\angle PDI=\alpha$ so $\angle QDI=\alpha$, also note that $\overline{PO}=\overline{QO}=\overline{CO}=d$. Because $I$ is the centroid of $ABC$, we know that ratio of $\overline{CI}$ to $\overline{DI}$ is $2:1$. Since we've scaled the figure down, the length of $CD$ is $3\sqrt{3}$, from this it's easy to know that $\overline{CI}=2\sqrt{3}$ and $\overline{DI}=\sqrt{3}$. The following two equations arise: \begin{align} \sqrt{3}\tan{\left(\alpha\right)}+\sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &= \sqrt{d^{2}-12} \end{align} Using trig identities for the tangent, we find that \begin{align*} \sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&=\sqrt{3}\left(\frac{\tan{\left(120^{\circ}\right)}+\tan{\left(\text{-}\alpha\right)}}{1-\tan{\left(120^{\circ}\right)}\tan{\left(\text{-}\alpha\right)}}\right) \\ &= \sqrt{3}\left(\frac{\text{-}\sqrt{3}+\tan{\left(\text{-}\alpha\right)}}{1+\sqrt{3}\tan{\left(\text{-}\alpha\right)}}\right) \\ &= \sqrt{3}\left(\frac{\text{-}\sqrt{3}-\tan{\left(\alpha\right)}}{1-\sqrt{3}\tan{\left(\alpha\right)}}\right) \\ &= \frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}.\end{align*} Okay, now we can plug this into $\left(1\right)$ to get: \begin{align}\sqrt{3}\tan{\left(\alpha\right)}+\frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}&=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &= \sqrt{d^{2}-12} \end{align} Notice that $\alpha$ only appears in the above system of equations in the form of $\sqrt{3}\tan{\left(\alpha\right)}$, we can set $\sqrt{3}\tan{\left(\alpha\right)}=a$ for convenience since we really only care about $d$. Now we have \begin{align}a+\frac{a+3}{a-1}&=2d \\ a - d &= \sqrt{d^{2}-12} \end{align} Looking at $\left(2\right)$, it's tempting to square it to get rid of the square-root so now we have: \begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\ a - 2ad &= \text{-}12 \end{align*} See the sneaky $2d$ in the above equation? That we means we can substitute it for $a+\frac{a+3}{a-1}$: \begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\ a^{2} - a\left(a+\frac{a+3}{a-1}\right) &= \text{-}12 \\ a^{2}-a^{2}-\frac{a^{2}+3a}{a-1} &=\text{-}12 \\ -\frac{a^{2}+3a}{a-1}&=\text{-}12 \\ \text{-}a^{2}-3a&=\text{-}12a+12 \\ 0 &= a^{2}-9a+12 \end{align*} Use the quadratic formula, we find that $a=\frac{9\pm\sqrt{9^{2}-4\left(1\right)\left(12\right)}}{2\left(1\right)}=\frac{9\pm\sqrt{33}}{2}$ - the two solutions were expected because $a$ can be $\angle PDI$ or $\angle QDI$. We can plug this into $\left(1\right)$: \begin{align*}a+\frac{a+3}{a-1}&=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{\frac{9\pm\sqrt{33}}{2}+3}{\frac{9\pm\sqrt{33}}{2}-1}=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{15\pm\sqrt{33}}{7\pm\sqrt{33}} &= 2d\end{align*} I'll use $a=\frac{9+\sqrt{33}}{2}$ because both values should give the same answer for $d$. \begin{align*} \frac{9+\sqrt{33}}{2}+\frac{15+\sqrt{33}}{7+\sqrt{33}} &= 2d \\ \frac{\left(9+\sqrt{33}\right)\left(7+\sqrt{33}\right)+\left(2\right)\left(15+\sqrt{33}\right)}{\left(2\right)\left(7+\sqrt{33}\right)} &= 2d \\ \frac{63+33+16\sqrt{33}+30+2\sqrt{33}}{14+2\sqrt{33}} &= 2d \\ \frac{126+18\sqrt{33}}{14+2\sqrt{33}} &= 2d \\ 9 &= 2d \\ \frac{9}{2} &= d\end{align*} Wait! Before you get excited, remember that we scaled the entire figure by $100$?? That means that the answer is $d=100\times\frac{9}{2}=\boxed{450}$. An alternate way of proceeding after finding $a$ (credit to riemanntensor), was to average the two possible values, you can see for yourself why this would work.

-fatant

See also

2016 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AIME Problems and Solutions

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