2020 AIME I Problems/Problem 15
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 in the form where and are positive integers, and is not divisible by the square of any prime. Find
Solution 1
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 1a
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.
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 .
Solution 3 (Official MAA 1)
Extend to intersect again at . The Power of a Point Theorem yields . Because , and and are both complements to , it follows that , implying that bisects , so the length of the altitude from to is .
Let the circumcircle of be . Because , the two triangles must have the same circumradius. Because the circumcircle of is , the circles and have the same radius . Denote the centers of and by and , respectively, and let be the midpoint of . Note that trapezoid has . Also and . Because is a translation of in the direction of , it follows that . Finally, the Pythagorean Theorem applied to yields . Let be the projection of onto . Then , so the Pythagorean Theorem applied to yields Solving for gives . It follows from properties of the orthocenter thatsoTherefore by the Extended Law of Sinesso The requested sum is .
Solution 4 (Official MAA 2)
Let be the intersection point of line and , noting that . Because the area of is , it suffices to compute and separately. As in the previous solution, . The value of can be found using the following lemma.
Lemma: Triangle is isosceles with base .
Proof: Because the circumcircle of , , and have the same radius, there exists a translation sending the former to the latter. Because is parallel to the line connecting the centers of the two circles, must send to , meaning also sends to the tangent to at . But this means that this tangent is parallel to , which implies the conclusion.
Applying Stewart's Theorem to yieldsimplying
By the Law of Cosines so Let be the radius of . By the Extended Law of Sines Then the solution proceeds as in Solution 3.
Solution 5 (Official MAA 3)
Define points and as above, and note that and . Let the circumcircle of be .
Extend past until it intersects line at point . Because line is a radical axis of and , it follows from the Power of a Point Theorem that from which . By Pythagorean Theorem .
Let and . By the Power of a Point Theorem On the other hand, from which . Therefore
Thus . Therefore as above.
Video Solution
https://www.youtube.com/watch?v=L7B20E95s4M
Solution 6
Let be circumcenter of let be circumradius of let be the image of circle over line (the circumcircle of ).
Let be the image of the reflection of over line lies on circle Let be the midpoint of Then lies on
(see here: https://brilliant.org/wiki/triangles-orthocenter/ or https://en.wikipedia.org/wiki/Altitude_(triangle) Russian)
lies on
lies on
We use properties of crossing chords and get We use properties of radius perpendicular chord and get We find We use properties of isosceles and find
We use and find
The area of vladimir.shelomovskii@gmail.com, vvsss
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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