2016 AIME II Problems/Problem 14

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 $\boxed{450}$ .

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 $\angle QDI=120^{\circ}-\alpha$ , 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 $d=100\times\frac{9}{2}=\boxed{450}$ . -fatant

Solution 4

We use the diagram from solution 3. From basic angle chasing, $180=\angle{QOC}+\angle{CO}P=2\angle{OCP}+2\angle{OCQ}=2\angle{QCP}$ so triangle QCP is a right triangle. This means that triangles $CQI$ and $CPI$ are similar. If we let $\angle{IDQ}=x$ and $\angle{PDI}=y$ , then we know $x+y=120$ and $\frac{PG}{GC}=\frac{GC}{GQ}\Rightarrow\frac{100\sqrt{3}\tan{y}}{200\sqrt{3}}=\frac{200\sqrt{3}}{100\sqrt{3}\tan{x}}\Rightarrow\tan{x}\tan{y}=4$ We also know that $PQ=2d=100\sqrt{3}(\tan{x}+\tan{y})$ $d=50\sqrt{3}(\tan{x}+\tan{y})$ $\frac{d}{1-\tan{x}\tan{y}}=50\sqrt{3}\cdot\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$ $\frac{d}{-3}=50\sqrt{3}\tan{(x+y)}$ $d=-150\sqrt{3}\tan{120}=-150\sqrt{3}(-\sqrt{3})=\boxed{450}$

-EZmath2006

Solution 5

We use the diagram from solution 3.

Let $BP = a$ and $BQ = b$ . Then, by Stewart's on $BPQ$ , we find $2x^3 + 2x^3 = a^2x + b^2x \implies a^2 + b^2 = 4x^2.$

The altitude from $P$ to $ABC$ is $\sqrt{a^2 - (200\sqrt{3})^2}$ so $PQ = 2x = \sqrt{a^2 - (200\sqrt{3})^2} + \sqrt{b^2 - (200\sqrt{3})^2}.$

Furthermore, the altitude from $P$ to $AB$ is $\sqrt{a^2 - 300^2}$ , so, by LoC and the dihedral condition, $a^2 - 300^2 + b^2 - 300^2 + \sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 4x^2.$

Squaring the equation for $PQ$ and substituting $a^2 + b^2 = 4x^2$ yields $2\sqrt{a^2 - (200\sqrt{3})^2}\sqrt{b^2 - (200\sqrt{3})^2} = 6\cdot 200^2.$

Substituting $a^2 + b^2 = 4x^2$ into the other equation, $\sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 2\cdot 300^2.$

Squaring both of these gives $a^2b^2-3\cdot 200^2(a^2 + b^2) + 9\cdot 200^4 = 9\cdot 200^4$ $a^2b^2 - 300^2(a^2+b^2) + 300^4 = 4\cdot 300^4.$

Substituting $a^2 + b^2 = 4x^2$ and solving for $x$ gives $\boxed{450}$ , as desired.

-mathtiger6

Solution 6 (Geometry)

Let $AB = a, M$ be midpoint $BC, I$ be the center of equilateral $\triangle ABC,$ $IM = b = \frac {a}{2\sqrt{3}}, O$ be the center of sphere $ABCPQ.$ Then $AI = 2b, AO = BO = PO =QO = d.$ $QA=QB=QC,PA=PB=PC \implies$ $POIQ\perp ABC, \angle PMQ = 120^\circ.$ (See upper diagram).

We construct the circle PQMD, use the formulas for intersecting chords and get $DI = 5b, FI = EO = \frac{3b}{2}$ $\implies FM = \frac{5b}{2}.$ (See lower diagram).

We apply the Law of Sine to $\triangle PMQ$ and get $2EM \sin 120^\circ =PQ$ $\implies r \sqrt{3} = 2d$ $\implies 3r^2 = 4d^2.$ We apply the Pythagorean Law on $\triangle AOI$ and $\triangle EFM$ and get $d^2 = 4b^2 + OI^2, r^2 = \frac {25b^2}{4} + EF^2 \implies$ $r = 3b\implies d = \frac {3a}{2} = \boxed {450}.$ vladimir.shelomovskii@gmail.com, vvsss

Solution 7

Let $M$ be the midpoint of $\overline{AB}$ and $X$ the center of $\triangle ABC$ . Then $P, O, Q, M, X, C$ all lie in the same vertical plane. We can make the following observations:

The equilateral triangle has side length $600$ , so $MC=300\sqrt{3}$ and $X$ divides $MC$ so that $MX=100\sqrt{3}$ and $XC=200\sqrt{3}$ ;
$O$ is the midpoint of $PQ$ since $O$ is equidistant from $A, B, C, P, Q$ – it is also the circumcenter of $\triangle PCQ$ ;
$\angle PMQ=120^{\circ}$ , the dihedral angle.

To make calculations easier, we will denote $100\sqrt{3}=m$ , so that $MX=m$ and $XC=2m$ .

$[asy] unitsize(20); pair P = (0, 12); pair Q = (0, -3); pair O = (P+Q)/2; pair M = (-3, 0); pair X = (0, 0); pair C = (6, 0); draw(P--O--Q); draw(M--X--C); draw(P--M--Q, blue); draw(Q--C--P); draw(circle((0, 4.5), 7.5)); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, E); dot(O); label("$M$", M, W); label("$X$", X, NE); label("$C$", C, E); label("$m$", (M+X)/2, N); label("$2m$", (X+C)/2, N); [/asy]$

Denote $PX=p$ and $QX=q$ , where the tangent addition formula on $\triangle PMQ$ yields $\frac{\tan\measuredangle PMX+\tan\measuredangle QMX}{1-\tan\measuredangle PMX\tan\measuredangle QMX}=\tan(120^{\circ})=-\sqrt{3}.$ Using $\tan\measuredangle PMX=\frac{p}{m}$ and $\tan\measuredangle QMX=\frac{q}{m}$ , we have $\frac{\frac{p}{m}+\frac{q}{m}}{1-\frac{p}{m}\cdot\frac{q}{m}}=-\sqrt{3}.$ After multiplying both numerator and denominator by $m^{2}$ we have $\frac{(p+q)m}{m^{2}-pq}=-\sqrt{3}.$ But note that $pq=(2m)(2m)=4m^{2}$ by power of a point at $X$ , where we deduce by symmetry that $MM^{\prime}=MX=m$ on the diagram below: $[asy] unitsize(20); pair P = (0, 12); pair Q = (0, -3); pair O = (P+Q)/2; pair M = (-3, 0); pair Mprime = (-6, 0); pair X = (0, 0); pair C = (6, 0); draw(P--O--Q); draw(Mprime--M--X--C); draw(P--M--Q, blue); draw(Q--C--P); draw(circle((0, 4.5), 7.5)); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, E); dot(O); label("$M$", M, S); label("$M^{\prime}$", Mprime, W); label("$X$", X, SE); label("$C$", C, E); label("$m$", (Mprime+M)/2, N); label("$m$", (M+X)/2, N); label("$2m$", (X+C)/2, N); [/asy]$

Thus $\begin{align*} \frac{(p+q)m}{m^{2}-4m^{2}}=-\sqrt{3} \\ \frac{p+q}{-3m}=-\sqrt{3} \\ p+q=\left(-\sqrt{3}\right)\left(-3m\right) \\ p+q=3\sqrt{3}\cdot m.\end{align*}$ Earlier we assigned the variable $m$ to the length $100\sqrt{3}$ which implies $PQ=\left(3\sqrt{3}\right)\left(100\sqrt{3}\right)=900$ . Thus the distance $d$ is equal to $\frac{PQ}{2}=\boxed{450}$ .

Solution 8 (Law of Cosines)

Let $Z$ be the center of $\triangle ABC$ . Let $A’$ be the midpoint of $BC$ . Let $ZA’ = c = 100\sqrt{3}$ and $ZA = 2c = 200\sqrt{3}$ . Let $PZ = a$ and $QZ = b$ .

$P$ , $Z$ , and $Q$ are collinear.

{incomplete; will finish soon} ~numerophile

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