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 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 2 (Short & Simple)

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 \[PQ=2d=100\sqrt{3}(\tan{x}+\tan{y})\] \[d=50\sqrt{3}(\tan{x}+\tan{y})\] \[\frac{d}{1-\tan{x}\tan{y}}=50\sqrt{3}\cdot\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}\] \[\frac{d}{-3}=50\sqrt{3}\tan{(x+y)}\] \[d=-150\sqrt{3}\tan{120}=-150\sqrt{3}(-\sqrt{3})=\boxed{450}\]


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.


Solution 6 (Geometry)

2016 AIME II 14.png
2016 AIME II 14a.png

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 $QZ = b$. We will be working in the plane that contains the points: $A$, $P$, $A’$, $Q$, $O$, and $Z$.

Since $P$, $O$, and $Q$ are collinear and $PO = QO = AO$, $\triangle PAQ$ is a right triangle with $\angle PAQ = 90^{\circ}$. Since $AZ \perp PQ$, $(PZ)(QZ) = (AZ)^2 = ab = (2c)^2 = 120000$.

$PA’ = \sqrt{a^2 + c^2}$, $QA’ = \sqrt{b^2 + c^2}$, $PQ = a + b$, and $\angle PAQ = 120^{\circ}$. By Law of Cosines \[(a + b)^2 = a^2 + b^2 + 2c^2 + \sqrt{a^2b^2 + a^2c^2 + b^2c^2 + c^4}\]. Substituting $4c^2$ for $ab$ and simplifying, we get \[6c = \sqrt{17c^2 + a^2 + b^2}\]. Squaring and simplifying, we get \[a^2 + b^2 = 19c^2 = 570000\]. Adding $2ab = 8c^2$ to both sides we get $PQ = a + b = 900$. Since $O$ is the midpoint of $PQ$, $d = PO = \boxed{450}$


Video Solution by MOP 2024



See also

2016 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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