Difference between revisions of "2016 AIME II Problems/Problem 14"
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==Solution== | ==Solution== | ||
− | 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 Shaddoll | Solution by Shaddoll |
Revision as of 10:55, 10 June 2016
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
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 by Shaddoll
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.