2020 USOJMO Problems/Problem 2
Contents
Problem
Let be the incircle of a fixed equilateral triangle . Let be a variable line that is tangent to and meets the interior of segments and at points and , respectively. A point is chosen such that and . Find all possible locations of the point , over all choices of .
Solution 1
Call a point good if it is a possible location for . Let the incircle of touch at , at , and at . Also, let the center of the incircle be . Clearly, a point is good iff it lies on the circle containing with center as well as the circle containing with center . Call these circles and , respectively.
Note that point can only lie on minor arc (excluding the endpoints).
Claim: A point is good iff and .
Proof: WLOG, let be on the same side of as . Then we have that are collinear. In particular, we have . Then , , , so that . Similarly, , so is good. Then , the reflection of across , is also good. But and have at most two intersections, so and must be these intersections, and since a point is good iff it lies on both circles, we are done.
Now, we know that are collinear. Then we have two cases:
Case 1: and lie on the same side of . Then we have , so that . Then we have that lie on a circle with center . Note that because lies on , must lie on . So, one of the solutions is .
Case 2: and lie on opposite sides of . Then extend out from to such that , and extend out from to such that . Then we see that , so that .
Solution 2
We claim that can lie on minor arc of the circumcircle of triangle ABC, and it can also lie on the dilation of this arc about the center of triangle with a factor of
Let and be the feet of the angle bisectors from points and respectively. Trivially, is also the medial triangle, orthic triangle, and contact triangle (ABC is equilateral).
Let be the incenter of ABC. Trivially, is the centroid, orthocenter, and circumcenter of ABC (ABC is equilateral). Also, where is the radius of circle (This is trivial). is the point of tangency of and segment .
R has to lie on the intersection of circles (center P, radius PA) and (center Q, radius QB), and for each choice of P, there exist two locations for R. The location that we claim to lie on the minor arc AB of the circumcircle of ABC shall be denoted M, and the other location shall be denoted N.
Define triangle XYZ to be the homothety of triangle ABC about I with a factor of -2.
Critical claim: M, T, I, and N are collinear.
Proof:
First we shall prove that T lies on MN using phantom points.
Let the intersection of MN and PQ be denoted as K. We shall prove that K and T are the same point.
Let and . Because of the equal tangent theorem, and . Hence, by the pythagorean theorem (recall ), and . Since PN = PA and QN = QB, then and .
PQ is the perpendicular bisector of MN because MN is the radical axis of and . Hence, M is the reflection of N across K. Also, NK is the altitude of triangle PNQ, so by using the pythagorean theorem and earlier expressions for and . However, . Now, we have a system of equations to solve for PK and QK in terms of p and q.
Dividing the first equation by the second (we can do this because p+q is always nonzero), we get . Combining this with our PK+QK result, we get and . However, and , and only one point can exist on PQ for which this result holds true. As a result, K and T are the same point, otherwise it is a contradiction. Hence M, T, and N are collinear.
. This is because both MN and IT are perpendicular to PQ (IT is perpendicular to PQ because PQ is a tangent with point of tangency T). However, both lines share point T, as discussed earlier. Hence, IT and MN are the same line, and M, T, I, and N are collinear.
In fact, from our earlier results from the lengths of PN and PT, we can use the pythagorean theorem to get that , a result that is always true and independent of P and Q! Also, because M is the reflection of N over K (which is the same as T), also. However, T varies based on P and Q. On the other hand, and M, T, I and N are collinear. Remembering our earlier definitions of M and N, we get that and , with M on the opposite side of N and T from I. Hence, M can be taken to N with a homothety about I with a factor of -2, and T can be taken to M with a homothety about I with a factor of -2. Since, trivially, the circumradius of ABC is 2r (ABC is equilateral), it seems like M can lie anywhere on the circumcircle of ABC.
However, we must take into account the restrictions on P and Q. This limits T to only minor arc DE on the incircle of ABC, hence, because of our earlier homothety statement, M is restricted to minor arc AB on the circumcircle of ABC. Because of our homothety statement about N, N has to lie on minor arc XY on the circumcircle of triangle XYZ.
Because we defined both M and N to be possible locations for R, .
-QED
-Solution by thanosaops
See Also
2020 USAJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
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