2020 AIME I Problems/Problem 15
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
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
As in the diagram, let ray extended hits BC at L and the circumcircle at say . By power of the point at H, we have . The three values we are given tells us that . L is the midpoint of (see here: https://www.cut-the-knot.org/Curriculum/Geometry/AltitudeAndCircumcircle.shtml ), so .
As in the diagram provided, let K be the intersection of and . By power of a point on the circumcircle of triangle , . By power of a point on the circumcircle of triangle , , thus . Solving gives or .
By the Pythagorean Theorem on triangle , . Now continue with solution 1.
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 .
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