2023 USAJMO Problems/Problem 6
Problem
Isosceles triangle , with , is inscribed in circle . Let be an arbitrary point inside such that . Ray intersects again at (other than ). Point (other than ) is chosen on such that . Line intersects rays and at points and , respectively. Prove that .
Solution
All angle and side length names are defined as in the figures below. Figure 1 is the diagram of the problem while Figure 2 is the diagram of the Ratio Lemma. Do note that the point names defined in the Ratio Lemma are not necessarily the same defined points on Figure 1.
First, we claim the Ratio Lemma:
We prove this as follows:
By the Law of Sines of we get: By the Law of Sines of we get: Dividing each of the corresponding equations, we get: Noting that we have:
We transition into our problem.
By Power of a Point on circle we have: By power of a point on circle we have: Using the transitive property, we have: Thus, by reverse Power of a Point, we have that quadrilaterals and are cyclic.
Now, we move to angle chasing.
- (from cyclic )
- (from the isosceles triangle)
- (from cyclic )
It is from here in which we focus on quadrilateral
Let and Now realize that all we want to show, as given by the problem statement, is
Thus,
We also have:
Calling back to what we want to show - if we want to show that it suffices to show that and But this is equivalent to showing
However, applying our Ratio Lemma to we have: Applying our Ratio Lemma to we have: From the transitive property, we have: Now, realize that we already have a shared side and equivalent angles Thus, we establish a stronger bound between these two triangles - a congruence.
Thus, these two triangles are similar, and our proof is complete.
mathboy282
(cred. to v_Enhance diagram)
See Also
2023 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
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