2022 USAJMO Problems/Problem 4
Problem
Let be a rhombus, and let and be points such that lies inside the rhombus, lies outside the rhombus, and . Prove that there exist points and on lines and such that is also a rhombus.
Solution
(Image of the solution is here [1])
Let's draw () perpendicular bisector of . Let be intersections of with and , respectively. is a kite. Let mid-point of . Let mid-point of (and also is mid-point of ). are on the line .
, , and so (side-side-side). By spiral similarity, . Hence, we get
Similarly, , , and so (side-side-side). From spiral similarity, . Thus,
If we can show that , then the kite will be a rhombus.
By spiral similarities, and . Then, .
. Then, . Also, in the right triangles and , . Therefore,
and we get .
(Lokman GÖKÇE)
See Also
2022 USAJMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
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