Difference between revisions of "2016 AIME II Problems/Problem 14"
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==Solution 1== | ==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>\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>. | 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>. | ||
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==Short Simple Solution== | ==Short Simple Solution== |
Revision as of 19:59, 18 October 2020
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
. An alternate way of proceeding after finding
(credit to riemanntensor), was to average the two possible values, you can see for yourself why this would work.
-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
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.