Difference between revisions of "2016 AIME II Problems/Problem 14"
(Solution) |
m (→Solution 3) |
||
(18 intermediate revisions by 12 users not shown) | |||
Line 1: | Line 1: | ||
+ | ==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>. | ||
− | ==Solution== | + | ==Solution 1== |
− | The inradius of <math>\triangle ABC</math> is <math>100\sqrt 3</math> and the circumradius is <math>200 \sqrt 3</math>. Now, consider the line perpendicular to plane <math>ABC</math> through the circumcenter of <math>\triangle ABC</math>. Note that <math>P,Q,O</math> must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since <math>P, Q, O</math> are collinear, and <math>OP=OQ</math>, we must have <math>O</math> is the midpoint of <math>PQ</math>. Now, Let <math>K</math> be the circumcenter of <math>\triangle ABC</math>, and <math>L</math> be the foot of the altitude from <math>A</math> to <math>BC</math>. We must have <math>\tan(\angle KLP+ \angle QLK)= \tan(120^{\circ})</math>. Setting <math>KP=x</math> and <math>KQ=y</math>, assuming WLOG <math>x>y</math>, we must have <math> | + | The inradius of <math>\triangle ABC</math> is <math>100\sqrt 3</math> and the circumradius is <math>200 \sqrt 3</math>. Now, consider the line perpendicular to plane <math>ABC</math> through the circumcenter of <math>\triangle ABC</math>. Note that <math>P,Q,O</math> must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since <math>P, Q, O</math> are collinear, and <math>OP=OQ</math>, we must have <math>O</math> is the midpoint of <math>PQ</math>. Now, Let <math>K</math> be the circumcenter of <math>\triangle ABC</math>, and <math>L</math> be the foot of the altitude from <math>A</math> to <math>BC</math>. We must have <math>\tan(\angle KLP+ \angle QLK)= \tan(120^{\circ})</math>. Setting <math>KP=x</math> and <math>KQ=y</math>, assuming WLOG <math>x>y</math>, we must have <math>\tan(120^{\circ})=-\sqrt{3}=\dfrac{\dfrac{x+y}{100 \sqrt{3}}}{\dfrac{30000-xy}{30000}}</math>. Therefore, we must have <math>100(x+y)=xy-30000</math>. Also, we must have <math>\left(\dfrac{x+y}{2}\right)^{2}=\left(\dfrac{x-y}{2}\right)^{2}+120000</math> by the Pythagorean theorem, so we have <math>xy=120000</math>, so substituting into the other equation we have <math>90000=100(x+y)</math>, or <math>x+y=900</math>. Since we want <math>\dfrac{x+y}{2}</math>, the desired answer is <math>\boxed{450}</math>. |
+ | |||
+ | ==Short Simple Solution== | ||
+ | Draw a good diagram. Draw <math>CH</math> as an altitude of the triangle. Scale everything down by a factor of <math>100\sqrt{3}</math>, so that <math>AB=2\sqrt{3}</math>. Finally, call the center of the triangle U. Draw a cross-section of the triangle via line <math>CH</math>, which of course includes <math>P, Q</math>. From there, we can call <math>OU=h</math>. There are two crucial equations we can thus generate. WLOG set <math>PU<QU</math>, then we call <math>PU=d-h, QU=d+h</math>. First equation: using the Pythagorean Theorem on <math>\triangle UOB</math>, <math>h^2+2^2=d^2</math>. Next, using the tangent addition formula on angles <math>\angle PHU, \angle UHQ</math> we see that after simplifying <math>-d^2+h^2=-4, 2d=3\sqrt{3}</math> in the numerator, so <math>d=\frac{3\sqrt{3}}{2}</math>. Multiply back the scalar and you get <math>\boxed{450}</math>. Not that hard, was it? | ||
+ | |||
+ | ==Solution 3== | ||
+ | To make numbers more feasible, we'll scale everything down by a factor of <math>100</math> so that <math>\overline{AB}=\overline{BC}=\overline{AC}=6</math>. We should also note that <math>P</math> and <math>Q</math> must lie on the line that is perpendicular to the plane of <math>ABC</math> and also passes through the circumcenter of <math>ABC</math> (due to <math>P</math> and <math>Q</math> being equidistant from <math>A</math>, <math>B</math>, <math>C</math>), let <math>D</math> be the altitude from <math>C</math> to <math>AB</math>. 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 <math>\angle PDI=\alpha</math> so <math>\angle QDI=120^{\circ}-\alpha</math>, also note that <math>\overline{PO}=\overline{QO}=\overline{CO}=d</math>. Because <math>I</math> is the centroid of <math>ABC</math>, we know that ratio of <math>\overline{CI}</math> to <math>\overline{DI}</math> is <math>2:1</math>. Since we've scaled the figure down, the length of <math>CD</math> is <math>3\sqrt{3}</math>, from this it's easy to know that <math>\overline{CI}=2\sqrt{3}</math> and <math>\overline{DI}=\sqrt{3}</math>. The following two equations arise: <cmath>\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}</cmath> Using trig identities for the tangent, we find that <cmath>\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*}</cmath> Okay, now we can plug this into <math>\left(1\right)</math> to get: <cmath>\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}</cmath> Notice that <math>\alpha</math> only appears in the above system of equations in the form of <math>\sqrt{3}\tan{\left(\alpha\right)}</math>, we can set <math>\sqrt{3}\tan{\left(\alpha\right)}=a</math> for convenience since we really only care about <math>d</math>. Now we have <cmath>\begin{align}a+\frac{a+3}{a-1}&=2d \\ a - d &= \sqrt{d^{2}-12} \end{align}</cmath> Looking at <math>\left(2\right)</math>, it's tempting to square it to get rid of the square-root so now we have: <cmath>\begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\ a - 2ad &= \text{-}12 \end{align*}</cmath> See the sneaky <math>2d</math> in the above equation? That we means we can substitute it for <math>a+\frac{a+3}{a-1}</math>: <cmath>\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*}</cmath> Use the quadratic formula, we find that <math>a=\frac{9\pm\sqrt{9^{2}-4\left(1\right)\left(12\right)}}{2\left(1\right)}=\frac{9\pm\sqrt{33}}{2}</math> - the two solutions were expected because <math>a</math> can be <math>\angle PDI</math> or <math>\angle QDI</math>. We can plug this into <math>\left(1\right)</math>: <cmath>\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*}</cmath> I'll use <math>a=\frac{9+\sqrt{33}}{2}</math> because both values should give the same answer for <math>d</math>. <cmath>\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*}</cmath> Wait! Before you get excited, remember that we scaled the entire figure by <math>100</math>?? That means that the answer is <math>d=100\times\frac{9}{2}=\boxed{450}</math>. | ||
+ | -fatant | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | We use the diagram from solution 3. From basic angle chasing, | ||
+ | <cmath>180=\angle{QOC}+\angle{CO}P=2\angle{OCP}+2\angle{OCQ}=2\angle{QCP}</cmath> | ||
+ | so triangle QCP is a right triangle. This means that triangles <math>CQI</math> and <math>CPI</math> are similar. If we let <math>\angle{IDQ}=x</math> and <math>\angle{PDI}=y</math>, then we know <math>x+y=120</math> and <cmath>\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</cmath> We also know that <cmath>PQ=2d=100\sqrt{3}(\tan{x}+\tan{y})</cmath> <cmath>d=50\sqrt{3}(\tan{x}+\tan{y})</cmath> <cmath>\frac{d}{1-\tan{x}\tan{y}}=50\sqrt{3}\cdot\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}</cmath> <cmath>\frac{d}{-3}=50\sqrt{3}\tan{(x+y)}</cmath> <cmath>d=-150\sqrt{3}\tan{120}=-150\sqrt{3}(-\sqrt{3})=\boxed{450}</cmath> | ||
+ | |||
+ | -EZmath2006 | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | We use the diagram from solution 3. | ||
+ | |||
+ | Let <math>BP = a</math> and <math>BQ = b</math>. Then, by Stewart's on <math>BPQ</math>, we find | ||
+ | <cmath>2x^3 + 2x^3 = a^2x + b^2x \implies a^2 + b^2 = 4x^2.</cmath> | ||
+ | |||
+ | The altitude from <math>P</math> to <math>ABC</math> is <math>\sqrt{a^2 - (200\sqrt{3})^2}</math> so | ||
+ | <cmath>PQ = 2x = \sqrt{a^2 - (200\sqrt{3})^2} + \sqrt{b^2 - (200\sqrt{3})^2}.</cmath> | ||
+ | |||
+ | Furthermore, the altitude from <math>P</math> to <math>AB</math> is <math>\sqrt{a^2 - 300^2}</math>, so, by LoC and the dihedral condition, | ||
+ | <cmath>a^2 - 300^2 + b^2 - 300^2 + \sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 4x^2.</cmath> | ||
+ | |||
+ | Squaring the equation for <math>PQ</math> and substituting <math>a^2 + b^2 = 4x^2</math> yields | ||
+ | <cmath>2\sqrt{a^2 - (200\sqrt{3})^2}\sqrt{b^2 - (200\sqrt{3})^2} = 6\cdot 200^2.</cmath> | ||
+ | |||
+ | Substituting <math>a^2 + b^2 = 4x^2</math> into the other equation, | ||
+ | <cmath>\sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 2\cdot 300^2.</cmath> | ||
+ | |||
+ | Squaring both of these gives | ||
+ | <cmath>a^2b^2-3\cdot 200^2(a^2 + b^2) + 9\cdot 200^4 = 9\cdot 200^4</cmath> | ||
+ | <cmath>a^2b^2 - 300^2(a^2+b^2) + 300^4 = 4\cdot 300^4.</cmath> | ||
+ | |||
+ | Substituting <math>a^2 + b^2 = 4x^2</math> and solving for <math>x</math> gives <math>\boxed{450}</math>, as desired. | ||
+ | |||
+ | -mathtiger6 | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2016|n=II|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Revision as of 22:48, 16 March 2021
Contents
Problem
Equilateral has side length . Points and lie outside the plane of and are on opposite sides of the plane. Furthermore, , and , and the planes of and form a dihedral angle (the angle between the two planes). There is a point whose distance from each of and is . Find .
Solution 1
The inradius of is and the circumradius is . Now, consider the line perpendicular to plane through the circumcenter of . Note that must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since are collinear, and , we must have is the midpoint of . Now, Let be the circumcenter of , and be the foot of the altitude from to . We must have . Setting and , assuming WLOG , we must have . Therefore, we must have . Also, we must have by the Pythagorean theorem, so we have , so substituting into the other equation we have , or . Since we want , the desired answer is .
Short Simple Solution
Draw a good diagram. Draw as an altitude of the triangle. Scale everything down by a factor of , so that . Finally, call the center of the triangle U. Draw a cross-section of the triangle via line , which of course includes . From there, we can call . There are two crucial equations we can thus generate. WLOG set , then we call . First equation: using the Pythagorean Theorem on , . Next, using the tangent addition formula on angles we see that after simplifying in the numerator, so . Multiply back the scalar and you get . Not that hard, was it?
Solution 3
To make numbers more feasible, we'll scale everything down by a factor of so that . We should also note that and must lie on the line that is perpendicular to the plane of and also passes through the circumcenter of (due to and being equidistant from , , ), let be the altitude from to . We can draw a vertical cross-section of the figure then: We let so , also note that . Because is the centroid of , we know that ratio of to is . Since we've scaled the figure down, the length of is , from this it's easy to know that and . The following two equations arise: Using trig identities for the tangent, we find that Okay, now we can plug this into to get: Notice that only appears in the above system of equations in the form of , we can set for convenience since we really only care about . Now we have Looking at , it's tempting to square it to get rid of the square-root so now we have: See the sneaky in the above equation? That we means we can substitute it for : Use the quadratic formula, we find that - the two solutions were expected because can be or . We can plug this into : I'll use because both values should give the same answer for . Wait! Before you get excited, remember that we scaled the entire figure by ?? That means that the answer is . -fatant
Solution 4
We use the diagram from solution 3. From basic angle chasing, so triangle QCP is a right triangle. This means that triangles and are similar. If we let and , then we know and We also know that
-EZmath2006
Solution 5
We use the diagram from solution 3.
Let and . Then, by Stewart's on , we find
The altitude from to is so
Furthermore, the altitude from to is , so, by LoC and the dihedral condition,
Squaring the equation for and substituting yields
Substituting into the other equation,
Squaring both of these gives
Substituting and solving for gives , as desired.
-mathtiger6
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.