Difference between revisions of "1997 USAMO Problems/Problem 2"
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Case 1: <math>D,E,F</math> are non-collinear | Case 1: <math>D,E,F</math> are non-collinear | ||
− | Observe that since <math>D,E,F</math> lie on perpendicular bisectors, then we get that <math>DC=DB</math>, <math>EC=EA</math>, and <math>FA=FB</math>. This motivates us to construct a circle <math>\omega_1</math> | + | Observe that since <math>D,E,F</math> lie on perpendicular bisectors, then we get that <math>DC=DB</math>, <math>EC=EA</math>, and <math>FA=FB</math>. This motivates us to construct a circle <math>\omega_1</math> centered at <math>D</math> with radius <math>DC</math>, and similarly construct <math>\omega_2</math> and <math>\omega_3</math> respectively for <math>E</math> and <math>F</math>. |
− | Now, clearly <math>\omega_1</math> and <math>\omega_2</math> intersect at <math>C</math> and some other point. Now, we know that <math>DE</math> is the line containing the two centers. So, the line perpendicular to <math>DE</math> and through <math>C</math> must be the radical axis, which is exactly the line that the problem describes! We do this similarly for the others pairs of circles. | + | Now, clearly <math>\omega_1</math> and <math>\omega_2</math> intersect at <math>C</math> and some other point. Now, we know that <math>DE</math> is the line containing the two centers. So, the line perpendicular to <math>DE</math> and passing through <math>C</math> must be the radical axis of <math>\omega_1</math> and <math>\omega_2</math>, which is exactly the line that the problem describes! We do this similarly for the others pairs of circles. |
Now by the radical lemma, the pairwise radical axes of <math>\omega_1,\omega_2,\omega_3</math> are concurrent, as desired, and they intersect at the radical center. | Now by the radical lemma, the pairwise radical axes of <math>\omega_1,\omega_2,\omega_3</math> are concurrent, as desired, and they intersect at the radical center. | ||
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Case 2: <math>D,E,F</math> are collinear | Case 2: <math>D,E,F</math> are collinear | ||
− | Now, we are drawing perpendicular lines from <math>A</math>, <math>B</math>, and <math>C</math> onto the single line <math>DEF</math>. Clearly, these lines are parallel and | + | Now, we are drawing perpendicular lines from <math>A</math>, <math>B</math>, and <math>C</math> onto the single line <math>DEF</math>. Clearly, these lines are parallel and concur at the point of infinity. |
+ | ==Solution 3== | ||
+ | |||
+ | Notice that <math>DEF</math> and <math>ABC</math> are orthologic, so the perpendiculars concur at the opposite center of orthology. | ||
==See Also== | ==See Also== | ||
{{USAMO newbox|year=1997|num-b=1|num-a=3}} | {{USAMO newbox|year=1997|num-b=1|num-a=3}} |
Latest revision as of 18:23, 24 July 2024
Contents
[hide]Problem
is a triangle. Take points on the perpendicular bisectors of respectively. Show that the lines through perpendicular to respectively are concurrent.
Solution 1
Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By Carnot's Theorem, The three lines are concurrent if
But this is clearly true, since D lies on the perpendicular bisector of BC, BD = DC.
QED
Solution 2
We split this into two cases:
Case 1: are non-collinear
Observe that since lie on perpendicular bisectors, then we get that , , and . This motivates us to construct a circle centered at with radius , and similarly construct and respectively for and .
Now, clearly and intersect at and some other point. Now, we know that is the line containing the two centers. So, the line perpendicular to and passing through must be the radical axis of and , which is exactly the line that the problem describes! We do this similarly for the others pairs of circles.
Now by the radical lemma, the pairwise radical axes of are concurrent, as desired, and they intersect at the radical center.
Case 2: are collinear
Now, we are drawing perpendicular lines from , , and onto the single line . Clearly, these lines are parallel and concur at the point of infinity.
Solution 3
Notice that and are orthologic, so the perpendiculars concur at the opposite center of orthology.
See Also
1997 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.