Difference between revisions of "1990 USAMO Problems/Problem 5"
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We know that <math>BCB'C'</math> is a cyclic quadrilateral. Hence, | We know that <math>BCB'C'</math> is a cyclic quadrilateral. Hence, | ||
− | < | + | <math>HB \cdot HB' = HC \cdot HC' </math> |
− | < | + | <math>\implies Pow_{\omega_{1}} = Pow_{\omega_{2}} </math> |
− | < | + | <math>\implies HM \cdot HN = HP \cdot HQ </math> |
− | < | + | <math>\implies MPNQ </math><math> is cyclic </math>\raggedright\blacksquare $. |
== See Also == | == See Also == |
Revision as of 00:42, 27 September 2024
Contents
[hide]Problem
An acute-angled triangle is given in the plane. The circle with diameter intersects altitude and its extension at points and , and the circle with diameter intersects altitude and its extensions at and . Prove that the points lie on a common circle.
Solution 1
Let be the intersection of the two circles (other than ). is perpendicular to both , implying , , are collinear. Since is the foot of the altitude from : , , are concurrent, where is the orthocentre.
Now, is also the intersection of , which means that , , are concurrent. Since , , , and , , , are cyclic, , , , are cyclic by the radical axis theorem.
Solution 2
Define as the foot of the altitude from to . Then, is the orthocenter. We will denote this point as . Since and are both , lies on the circles with diameters and .
Now we use the Power of a Point theorem with respect to point . From the circle with diameter we get . From the circle with diameter we get . Thus, we conclude that , which implies that , , , and all lie on a circle.
Solution 3 (Radical Lemma)
Let be the circumcircle with diameter and be the circumcircle with diameter . We claim that the second intersection of and other than is , where is the feet of the perpendicular from to segment . Note that so lies on Similarly, lies on . Hence, is the radical axis of and . By the Radical Lemma, it suffices to prove that the intersection of lines and lie on . But, is the same line as and is the same line as . Since , and intersect at the orthocenter , lies on the radical axis and we are done.
Solution 4
We know that is a cyclic quadrilateral. Hence,
\raggedright\blacksquare $.
See Also
1990 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.