Difference between revisions of "2024 USAJMO Problems/Problem 1"
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+ | ==Solution 4 (Coord Bash)== | ||
+ | [Will add when have time] | ||
+ | ~KevinChen_Yay | ||
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Revision as of 09:38, 24 April 2024
Problem
Let be a cyclic quadrilateral with and . Points and are selected on segment such that . Points and are selected on segment such that . Prove that is a cyclic quadrilateral.
Solution 1
First, let and be the midpoints of and , respectively. It is clear that , , , and . Also, let be the circumcenter of .
By properties of cyclic quadrilaterals, we know that the circumcenter of a cyclic quadrilateral is the intersection of its sides' perpendicular bisectors. This implies that and . Since and are also bisectors of and , respectively, if is indeed a cyclic quadrilateral, then its circumcenter is also at . Thus, it suffices to show that .
Notice that , , and . By SAS congruency, . Similarly, we find that and . We now need only to show that these two pairs are equal to each other.
Draw the segments connecting to , , , and .
Also, let be the circumradius of . This means that . Recall that and . Notice the several right triangles in our figure.
Let us apply Pythagorean Theorem on . We can see that
Let us again apply Pythagorean Theorem on . We can see that
Let us apply Pythagorean Theorem on . We get .
We finally apply Pythagorean Theorem on . This becomes .
This is the same expression as we got for . Thus, , and recalling that and , we have shown that . We are done. QED
~Technodoggo
Solution 2
We can consider two cases: or The first case is trivial, as and we are done due to symmetry. For the second case, WLOG, assume that and are located on and respectively. Extend and to a point and by Power of a Point, we have which may be written as or We can translate this to so and therefore by the Converse of Power of a Point is cyclic, and we are done.
Solution 3
All 4 corners of have equal power of a point () with respect to the circle , with center .
Draw diameters (of length ) of circle through and , with length . Let be the distance from to the circle along a diameter, and likewise be distance from to the circle.
Then and (radius). Therefore, and . But , , and by symmetry around the perpendicular bisectors of and , so are all equidistant from , forming a circumcircle around .
-BraveCobra22aops and oinava
Solution 4 (Coord Bash)
[Will add when have time] ~KevinChen_Yay
See Also
2024 USAJMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.