Difference between revisions of "1990 USAMO Problems/Problem 5"
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Now, <math>H</math> is also the intersection of <math>BB'</math>, <math>CC'</math> which means that <math>AA'</math>, <math>MN</math>, <math>PQ</math> are concurrent. Since <math>A</math>, <math>M</math>, <math>N</math>, <math>A'</math> and <math>A</math>, <math>P</math>, <math>Q</math>, <math>A'</math> are cyclic, <math>M</math>, <math>N</math>, <math>P</math>, <math>Q</math> are cyclic by the radical axis theorem. | Now, <math>H</math> is also the intersection of <math>BB'</math>, <math>CC'</math> which means that <math>AA'</math>, <math>MN</math>, <math>PQ</math> are concurrent. Since <math>A</math>, <math>M</math>, <math>N</math>, <math>A'</math> and <math>A</math>, <math>P</math>, <math>Q</math>, <math>A'</math> are cyclic, <math>M</math>, <math>N</math>, <math>P</math>, <math>Q</math> are cyclic by the radical axis theorem. | ||
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+ | == Solution 2 == | ||
+ | |||
+ | Define <math>A'</math> as the foot of the altitude from <math>A</math> to <math>BC</math>. Then, <math>AA' \cap BB' \cap CC'</math> is the orthocenter. We will denote this point as <math>H</math> | ||
+ | Since <math>\angle AA'C</math> and <math>\angle AA'B</math> are both <math>90^{\Circle}</math>, <math>A'</math> lies on the circles with diameters <math>AC</math> and <math>AB</math>. | ||
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+ | Now we use the Power of a Point theorem with respect to point <math>H</math>. From the circle with diameter <math>AB</math> we get <math>AH \cdot A'H = MH \cdot NH</math>. From the circle with diameter <math>AC</math> we get <math>AH \cdot A'H = PH \cdot QH</math>. Thus, we conclude that <math>PH \cdot QH = MH \cdot NH</math>, which implies that <math>P</math>, <math>Q</math>, <math>M</math>, and <math>N</math> all lie on a circle. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 16:32, 19 July 2017
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
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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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.