Difference between revisions of "2024 USAJMO Problems/Problem 1"
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==Problem== | ==Problem== | ||
− | + | Let <math>ABCD</math> be a cyclic quadrilateral with <math>AB = 7</math> and <math>CD = 8</math>. Points <math>P</math> and <math>Q</math> are selected on segment <math>AB</math> such that <math>AP = BQ = 3</math>. Points <math>R</math> and <math>S</math> are selected on segment <math>CD</math> such that <math>CR = DS = 2</math>. Prove that <math>PQRS</math> is a cyclic quadrilateral. | |
==Solution 1== | ==Solution 1== |
Revision as of 00:31, 22 March 2024
Contents
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
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.