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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.
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==Solution==
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==Solution 1==
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Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By Carnot's Thereom, The three lines are concurrent if <cmath>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</cmath> But this is clearly true, because, since D lies on the perpendicular bisector of BC, BD = DC.
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Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By [[Carnot's Theorem]], The three lines are [[concurrent]] if
  
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QED
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<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>
  
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{{USAMO box|year=1997|num-b=1|num-a=3}}
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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
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 +
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.
 +
 
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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==
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Notice that <math>DEF</math> and <math>ABC</math> are orthologic, so the perpendiculars concur at the opposite center of orthology.
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==See Also==
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{{USAMO newbox|year=1997|num-b=1|num-a=3}}
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[[Category:Olympiad Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 18:23, 24 July 2024

Problem

$\triangle ABC$ is a triangle. Take points $D, E, F$ on the perpendicular bisectors of $BC, CA, AB$ respectively. Show that the lines through $A, B, C$ perpendicular to $EF, FD, DE$ 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

$FA'^2-EA'^2+EC'^2-DC'^2+DB'^2-FB'^2 = AF^2-AE^2+CE^2-CD^2+BD^2-BF^2 = 0$

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: $D,E,F$ are non-collinear

Observe that since $D,E,F$ lie on perpendicular bisectors, then we get that $DC=DB$, $EC=EA$, and $FA=FB$. This motivates us to construct a circle $\omega_1$ centered at $D$ with radius $DC$, and similarly construct $\omega_2$ and $\omega_3$ respectively for $E$ and $F$.

Now, clearly $\omega_1$ and $\omega_2$ intersect at $C$ and some other point. Now, we know that $DE$ is the line containing the two centers. So, the line perpendicular to $DE$ and passing through $C$ must be the radical axis of $\omega_1$ and $\omega_2$, 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 $\omega_1,\omega_2,\omega_3$ are concurrent, as desired, and they intersect at the radical center.

Case 2: $D,E,F$ are collinear

Now, we are drawing perpendicular lines from $A$, $B$, and $C$ onto the single line $DEF$. Clearly, these lines are parallel and concur at the point of infinity.

Solution 3

Notice that $DEF$ and $ABC$ 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

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