Difference between revisions of "2024 USAJMO Problems/Problem 1"
Virjoy2001 (talk | contribs) (→Solution 1) |
Virjoy2001 (talk | contribs) m (→Solution 2) |
||
Line 119: | Line 119: | ||
==Solution 2== | ==Solution 2== | ||
− | We can consider two cases: <math>AB \parallel CD</math> or <math>AB \nparallel CD.</math> The first case is trivial, as <math>PQ \parallel RS</math> and we are done. For the second case, WLOG, assume that <math>A</math> and <math>C</math> are located on <math>XB</math> and <math>XD</math> respectively. Extend <math>AB</math> and <math>CD</math> to a point <math>X,</math> and by Power of a Point, we have <cmath>XA\cdot XB = XC \cdot XD,</cmath> which may be written as <cmath>XA \cdot (XA+7) = XC \cdot (XC+8),</cmath> or <cmath>XA^2 + 7XA = XC^2 + 8XC.</cmath> We can translate this to <cmath>XA^2 + 7XA +12 = XC^2 + 8XC +12,</cmath> so <cmath>XP\cdot XQ = (XA+3)(XA+4)=(XC+2)(XC+6)= XR\cdot XS,</cmath> and therefore by the Converse of Power of a Point <math>PQRS</math> is cyclic, and we are done. | + | We can consider two cases: <math>AB \parallel CD</math> or <math>AB \nparallel CD.</math> The first case is trivial, as <math>PQ \parallel RS</math> and we are done due to symmetry. For the second case, WLOG, assume that <math>A</math> and <math>C</math> are located on <math>XB</math> and <math>XD</math> respectively. Extend <math>AB</math> and <math>CD</math> to a point <math>X,</math> and by Power of a Point, we have <cmath>XA\cdot XB = XC \cdot XD,</cmath> which may be written as <cmath>XA \cdot (XA+7) = XC \cdot (XC+8),</cmath> or <cmath>XA^2 + 7XA = XC^2 + 8XC.</cmath> We can translate this to <cmath>XA^2 + 7XA +12 = XC^2 + 8XC +12,</cmath> so <cmath>XP\cdot XQ = (XA+3)(XA+4)=(XC+2)(XC+6)= XR\cdot XS,</cmath> and therefore by the Converse of Power of a Point <math>PQRS</math> is cyclic, and we are done. |
- [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8] | - [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8] |
Revision as of 18:46, 24 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
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.
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.