Difference between revisions of "1997 USAMO Problems/Problem 2"
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==Problem== | ==Problem== | ||
<math>\triangle ABC</math> is a triangle. Take points <math>D, E, F</math> on the perpendicular bisectors of <math>BC, CA, AB</math> respectively. Show that the lines through <math>A, B, C</math> perpendicular to <math>EF, FD, DE</math> respectively are concurrent. | <math>\triangle ABC</math> is a triangle. Take points <math>D, E, F</math> on the perpendicular bisectors of <math>BC, CA, AB</math> respectively. Show that the lines through <math>A, B, C</math> perpendicular to <math>EF, FD, DE</math> respectively are concurrent. | ||
| − | ==Solution== | + | ==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 | |
| − | {{USAMO | + | <center><math>FA'^2-EA'^2+EC'^2-DC'^2+DB'^2-FB'^2 = AF^2-AE^2+CE^2-CD^2+BD^2-BF^2 = 0</math></center> |
| + | |||
| + | 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: <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> 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 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. | ||
| + | |||
| + | 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 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== | ||
| + | {{USAMO newbox|year=1997|num-b=1|num-a=3}} | ||
| + | |||
| + | [[Category:Olympiad Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:23, 24 July 2024
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. 
