Difference between revisions of "2016 AIME II Problems/Problem 14"
Mathgeek2006 (talk | contribs) m |
Ninjaforce (talk | contribs) |
||
(29 intermediate revisions by 16 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>. |
− | Solution by | + | ==Solution 2 (Short & Simple)== |
+ | 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> | ||
+ | -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 | ||
+ | |||
+ | ==Solution 6 (Geometry)== | ||
+ | [[File:2016 AIME II 14.png|400px|right]] | ||
+ | [[File:2016 AIME II 14a.png|400px|right]] | ||
+ | Let <math>AB = a, M</math> be midpoint <math>BC, I</math> be the center of equilateral <math>\triangle ABC,</math> | ||
+ | <math>IM = b = \frac {a}{2\sqrt{3}}, O</math> be the center of sphere <math>ABCPQ.</math> | ||
+ | Then <cmath>AI = 2b, AO = BO = PO =QO = d.</cmath> | ||
+ | <cmath>QA=QB=QC,PA=PB=PC \implies</cmath> | ||
+ | <cmath>POIQ\perp ABC, \angle PMQ = 120^\circ.</cmath> | ||
+ | (See upper diagram). | ||
+ | |||
+ | We construct the circle PQMD, use the formulas for intersecting chords and get | ||
+ | <cmath>DI = 5b, FI = EO = \frac{3b}{2}</cmath> | ||
+ | <cmath>\implies FM = \frac{5b}{2}.</cmath> | ||
+ | (See lower diagram). | ||
+ | |||
+ | We apply the Law of Sine to <math>\triangle PMQ</math> and get | ||
+ | <cmath>2EM \sin 120^\circ =PQ</cmath> | ||
+ | <cmath>\implies r \sqrt{3} = 2d</cmath> | ||
+ | <cmath>\implies 3r^2 = 4d^2.</cmath> | ||
+ | We apply the Pythagorean Law on <math>\triangle AOI</math> and <math>\triangle EFM</math> and get | ||
+ | <cmath>d^2 = 4b^2 + OI^2, r^2 = \frac {25b^2}{4} + EF^2 \implies</cmath> | ||
+ | <cmath>r = 3b\implies d = \frac {3a}{2} = \boxed {450}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 7== | ||
+ | Let <math>M</math> be the midpoint of <math>\overline{AB}</math> and <math>X</math> the center of <math>\triangle ABC</math>. Then <cmath>P, O, Q, M, X, C</cmath> all lie in the same vertical plane. We can make the following observations: | ||
+ | * The equilateral triangle has side length <math>600</math>, so <math>MC=300\sqrt{3}</math> and <math>X</math> divides <math>MC</math> so that <math>MX=100\sqrt{3}</math> and <math>XC=200\sqrt{3}</math>; | ||
+ | * <math>O</math> is the midpoint of <math>PQ</math> since <math>O</math> is equidistant from <math>A, B, C, P, Q</math> – it is also the circumcenter of <math>\triangle PCQ</math>; | ||
+ | * <math>\angle PMQ=120^{\circ}</math>, the dihedral angle. | ||
+ | |||
+ | To make calculations easier, we will denote <math>100\sqrt{3}=m</math>, so that <math>MX=m</math> and <math>XC=2m</math>. | ||
+ | |||
+ | <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 <math>PX=p</math> and <math>QX=q</math>, where the tangent addition formula on <math>\triangle PMQ</math> yields <cmath>\frac{\tan\measuredangle PMX+\tan\measuredangle QMX}{1-\tan\measuredangle PMX\tan\measuredangle QMX}=\tan(120^{\circ})=-\sqrt{3}.</cmath> Using <math>\tan\measuredangle PMX=\frac{p}{m}</math> and <math>\tan\measuredangle QMX=\frac{q}{m}</math>, we have <cmath>\frac{\frac{p}{m}+\frac{q}{m}}{1-\frac{p}{m}\cdot\frac{q}{m}}=-\sqrt{3}.</cmath> After multiplying both numerator and denominator by <math>m^{2}</math> we have <cmath>\frac{(p+q)m}{m^{2}-pq}=-\sqrt{3}.</cmath> But note that <math>pq=(2m)(2m)=4m^{2}</math> by power of a point at <math>X</math>, where we deduce by symmetry that <math>MM^{\prime}=MX=m</math> 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 <cmath> | ||
+ | |||
+ | ==Solution 8 (Law of Cosines)== | ||
+ | |||
+ | Let <math>Z</math> be the center of <math>\triangle ABC</math>. Let <math>A’</math> be the midpoint of <math>BC</math>. Let <math>ZA’ = c = 100\sqrt{3}</math> and <math>ZA = 2c = 200\sqrt{3}</math>. Let <math>PZ = a</math> and <math>QZ = b</math>. We will be working in the plane that contains the points: <math>A</math>, <math>P</math>, <math>A’</math>, <math>Q</math>, <math>O</math>, and <math>Z</math>. | ||
+ | |||
+ | Since <math>P</math>, <math>O</math>, and <math>Q</math> are collinear and <math>PO = QO = AO</math>, <math>\triangle PAQ</math> is a right triangle with <math>\angle PAQ = 90^{\circ}</math>. Since <math>AZ \perp PQ</math>, <math>(PZ)(QZ) = (AZ)^2 = ab = (2c)^2 = 120000</math>. | ||
+ | |||
+ | <math>PA’ = \sqrt{a^2 + c^2}</math>, <math>QA’ = \sqrt{b^2 + c^2}</math>, <math>PQ = a + b</math>, and <math>\angle PAQ = 120^{\circ}</math>. By Law of Cosines <cmath>(a + b)^2 = a^2 + b^2 + 2c^2 + \sqrt{a^2b^2 + a^2c^2 + b^2c^2 + c^4}</cmath>. Substituting <math>4c^2</math> for <math>ab</math> and simplifying, we get <cmath>6c = \sqrt{17c^2 + a^2 + b^2}</cmath>. Squaring and simplifying, we get <cmath>a^2 + b^2 = 19c^2 = 570000</cmath>. Adding <math>2ab = 8c^2</math> to both sides we get <math>PQ = a + b = 900</math>. Since <math>O</math> is the midpoint of <math>PQ</math>, <math>d = PO = \boxed{450}</math> | ||
+ | |||
+ | ~numerophile | ||
+ | |||
+ | ==Solution 9 (Coordinate Bash)== | ||
+ | |||
+ | Set <math>AB = s, M</math> as midpoint <math>BC, I</math> as the center of equilateral <math>\triangle ABC,</math> and by <math>30-60-90</math> triangle formulas, we know <math>IM = b = \frac {s}{2\sqrt{3}}, IA = IB = IC = \frac{s}{\sqrt{3}}</math>. | ||
+ | |||
+ | First note that <math>P</math> and <math>Q</math> are on the line perpendicular to plane <math>ABC</math> through the circumcenter of <math>\triangle ABC</math>. Then notice that point <math>O</math> is the midpoint of <math>PQ</math>, so it also lies on this axis. Then, we see that the position of <math>O</math> relative to the triangle fully determines the positions of both <math>P</math> and <math>Q</math>, so setting the coordinates of <math>O</math> as <math>(0,0,h)</math> where the origin is defined the circumcenter and <math>\triangle ABC</math> lies in the <math>xy</math>-plane, we get: <cmath> | ||
+ | |||
+ | Thus, the coordinates of <math>P</math> and <math>Q</math> are <math>(0,0,h\pm\sqrt{h^2+\frac{s^2}{3}})</math>. Now we make use of the angle condition. WLOG set the coordinates of <math>M</math> as <math>(\frac{s}{2\sqrt{3}},0,0)</math>. We know the angle between <math>PM</math> and <math>QM</math> is <math>120^{\circ}</math>, so after solving for the vectors, taking their dot product, equating it to <math>PM \cdot QM \cdot\cos{120^{\circ}}</math>, and <math>\textit{finally}</math> solving for <math>h</math> in terms of <math>s</math>, we get <cmath>\[ | ||
+ | h^2 = \frac{11}{48}s^2 \ | ||
+ | \implies OA^2 = h^2 + \frac{s^2}{3} = \frac{27}{48}s^2 \ | ||
+ | \implies OA = \frac{3}{4}s = \boxed{450} | ||
+ | \]</cmath> | ||
+ | |||
+ | |||
+ | (If someone feels like it, please feel free to fill in the rest of the details!) | ||
+ | |||
+ | ~ninjaforce | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/hyhIlsAR2hs | ||
+ | |||
+ | ~r00tsOfUnity | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=13|num-a=15}} | {{AIME box|year=2016|n=II|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 18:39, 25 November 2024
Contents
[hide]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
.
Solution 2 (Short & Simple)
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
Solution 6 (Geometry)
Let be midpoint
be the center of equilateral
be the center of sphere
Then
(See upper diagram).
We construct the circle PQMD, use the formulas for intersecting chords and get
(See lower diagram).
We apply the Law of Sine to and get
We apply the Pythagorean Law on
and
and get
vladimir.shelomovskii@gmail.com, vvsss
Solution 7
Let be the midpoint of
and
the center of
. Then
all lie in the same vertical plane. We can make the following observations:
- The equilateral triangle has side length
, so
and
divides
so that
and
;
is the midpoint of
since
is equidistant from
– it is also the circumcenter of
;
, the dihedral angle.
To make calculations easier, we will denote , so that
and
.
Denote and
, where the tangent addition formula on
yields
Using
and
, we have
After multiplying both numerator and denominator by
we have
But note that
by power of a point at
, where we deduce by symmetry that
on the diagram below:
Thus Earlier we assigned the variable
to the length
which implies
. Thus the distance
is equal to
.
Solution 8 (Law of Cosines)
Let be the center of
. Let
be the midpoint of
. Let
and
. Let
and
. We will be working in the plane that contains the points:
,
,
,
,
, and
.
Since ,
, and
are collinear and
,
is a right triangle with
. Since
,
.
,
,
, and
. By Law of Cosines
. Substituting
for
and simplifying, we get
. Squaring and simplifying, we get
. Adding
to both sides we get
. Since
is the midpoint of
,
~numerophile
Solution 9 (Coordinate Bash)
Set as midpoint
as the center of equilateral
and by
triangle formulas, we know
.
First note that and
are on the line perpendicular to plane
through the circumcenter of
. Then notice that point
is the midpoint of
, so it also lies on this axis. Then, we see that the position of
relative to the triangle fully determines the positions of both
and
, so setting the coordinates of
as
where the origin is defined the circumcenter and
lies in the
-plane, we get:
Thus, the coordinates of and
are
. Now we make use of the angle condition. WLOG set the coordinates of
as
. We know the angle between
and
is
, so after solving for the vectors, taking their dot product, equating it to
, and
solving for
in terms of
, we get
(If someone feels like it, please feel free to fill in the rest of the details!)
~ninjaforce
Video Solution by MOP 2024
~r00tsOfUnity
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.