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 18:19, 12 March 2020
Contents
[hide]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 | ||
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