Difference between revisions of "2020 AIME I Problems/Problem 15"
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== Solution 2 == | == Solution 2 == | ||
+ | <asy> | ||
+ | size(10cm); | ||
+ | pair A, B, C, D, H, K, O, P, L, M, X, Y; | ||
+ | A = (-15, 27); | ||
+ | B = (-24, 0); | ||
+ | C = (24, 0); | ||
+ | D = (-8.28, 18.04); | ||
+ | O = (0, 7); | ||
+ | P = (0, -7); | ||
+ | H = (-15, 13); | ||
+ | K = (-15, -13); | ||
+ | M = (0, 0); | ||
+ | L = (-15, 0); | ||
+ | X = (-24.9569, 5.53234); | ||
+ | Y = (8.39688, 30.5477); | ||
+ | draw(circle(O, 25)); | ||
+ | draw(circle(P, 25)); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(H -- K); | ||
+ | draw(A -- O -- P -- H -- cycle); | ||
+ | draw(X -- Y); | ||
+ | draw(O -- X, dashed); | ||
+ | draw(O -- Y, dashed); | ||
+ | draw(O -- B, dashed); | ||
+ | draw(O -- C, dashed); | ||
+ | |||
+ | label("$O$", O, ENE); | ||
+ | label("$A$", A, NW); | ||
+ | label("$B$", B, W); | ||
+ | label("$C$", C, E); | ||
+ | label("$H$", H, E); | ||
+ | label("$H'$", K, NE); | ||
+ | label("$X$", X, W); | ||
+ | label("$Y$", Y, NE); | ||
+ | label("$O'$", P, E); | ||
+ | label("$M$", M, NE); | ||
+ | label("$L$", L, NE); | ||
+ | label("$D$", D, NNE); | ||
+ | |||
+ | label("$2$", X -- H, NW); | ||
+ | label("$3$", H -- A, SW); | ||
+ | label("$6$", H -- Y, NW); | ||
+ | label("$R$", O -- Y, E); | ||
+ | |||
+ | dot(O); | ||
+ | dot(P); | ||
+ | dot(D); | ||
+ | dot(H); | ||
+ | |||
+ | </asy> | ||
+ | Diagram not to scale. | ||
+ | |||
+ | |||
+ | We first observe that <math>H'</math>, the image of the reflection of <math>H</math> over line <math>BC</math>, lies on circle <math>O</math>. This is because <math>\angle HBC = 90 - \angle C = \angle H'AC = \angle H'BC</math>. This is a well known lemma. The result of this observation is that circle <math>O'</math>, the circumcircle of <math>\triangle BHC</math> is the image of circle <math>O</math> over line <math>BC</math>, which in turn implies that <math>\overline{AH} = \overline{OO'}</math> and thus <math>AHO'O</math> is a parallelogram. That <math>AHO'O</math> is a parallelogram implies that <math>AO</math> is perpendicular to <math>\overline{XY}</math>, and thus divides segment <math>\overline{XY}</math> in two equal pieces, <math>\overline{XD}</math> and <math>\overline{DY}</math>, of length <math>4</math>. | ||
+ | |||
+ | |||
+ | Using Power of a Point, | ||
+ | <cmath>\overline{AH} \cdot \overline{HH'} = \overline{XH} \cdot \overline{HY} \Longrightarrow 3 \cdot \overline{HH'} = 2 \cdot 6 \Longrightarrow \overline{HH'} = 4</cmath> | ||
+ | This means that <math>\overline{HL} = \frac12 \cdot 4 = 2</math> and <math>\overline{AL} = 2 + 3 = 5</math>, where <math>L</math> is the foot of the altitude from <math>A</math> onto <math>BC</math>. All that remains to be found is the length of segment <math>\overline{BC}</math>. | ||
+ | |||
+ | Looking at right triangle <math>\triangle AHD</math>, we find that | ||
+ | <cmath>\overline{AD} = \sqrt{\overline{AH}^2 - \overline{HD}^2} = \sqrt{3^2 - 2^2} = \sqrt{5}</cmath> | ||
+ | Looking at right triangle <math>\triangle ODY</math>, we get the equation | ||
+ | <cmath>\overline{OY}^2 - \overline{HY}^2 = \overline{OD}^2 = \left(\overline{AO} - \overline{AD}\right)^2</cmath> | ||
+ | Plugging in known values, and letting <math>R</math> be the radius of the circle, we find that | ||
+ | <cmath>R^2 - 16 = (R - \sqrt{5})^2 = R^2 - 2\sqrt5 R + 5 \Longrightarrow R = \frac{21\sqrt5}{10}</cmath> | ||
+ | |||
+ | Recall that <math>AHO'O</math> is a parallelogram, so <math>\overline{AH} = \overline{OO'} = 3</math>. So, <math>\overline{OM} = \frac32</math>, where <math>M</math> is the midpoint of <math>\overline{BC}</math>. This means that | ||
+ | <cmath>\overline{BC} = 2\overline{BM} = 2\sqrt{R^2 - \left(\frac32\right)^2} = 2\sqrt{\frac{441}{20} - \frac{9}{4}} = \frac{6\sqrt{55}}{5}</cmath> | ||
+ | |||
+ | Thus, the area of triangle <math>\triangle ABC</math> is | ||
+ | <cmath>\frac{\overline{AL} \cdot \overline{BC}}{2} = \frac{5 \cdot \frac{6}{\sqrt{55}}{5}}{2} = \boxed{3\sqrt{55}}</cmath> | ||
+ | The answer is <math>3 + 55 = \boxed{058}</math>. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=I|num-b=14|after=Last Problem}} | {{AIME box|year=2020|n=I|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:19, 12 March 2020
Contents
Problem
Let be an acute triangle with circumcircle
and let
be the intersection of the altitudes of
Suppose the tangent to the circumcircle of
at
intersects
at points
and
with
and
The area of
can be written as
where
and
are positive integers, and
is not divisible by the square of any prime. Find
Solution
The following is a power of a point solution to this menace of a problem:
Let points be what they appear as in the diagram below. Note that is not insignificant; from here, we set
by PoP and trivial construction. Now,
is the reflection of
over
. Note
, and therefore by Pythagorean theorem we have
. Consider
. We have that
, and therefore we are ready to PoP with respect to
. Setting
, we obtain
by PoP on
, and furthermore, we have
. Now, we get
, and from
we take
However, squaring and manipulating with
yields that
and from here, since
we get the area to be
. ~awang11's sol
Solution 2
Diagram not to scale.
We first observe that , the image of the reflection of
over line
, lies on circle
. This is because
. This is a well known lemma. The result of this observation is that circle
, the circumcircle of
is the image of circle
over line
, which in turn implies that
and thus
is a parallelogram. That
is a parallelogram implies that
is perpendicular to
, and thus divides segment
in two equal pieces,
and
, of length
.
Using Power of a Point,
This means that
and
, where
is the foot of the altitude from
onto
. All that remains to be found is the length of segment
.
Looking at right triangle , we find that
Looking at right triangle
, we get the equation
Plugging in known values, and letting
be the radius of the circle, we find that
Recall that is a parallelogram, so
. So,
, where
is the midpoint of
. This means that
Thus, the area of triangle is
The answer is
.
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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.