Difference between revisions of "2013 USAMO Problems/Problem 6"
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<cmath>\left(x - \frac{ab}{b+c}\right)\left(x - \frac{ac}{b+c}\right) = 0.</cmath> | <cmath>\left(x - \frac{ab}{b+c}\right)\left(x - \frac{ac}{b+c}\right) = 0.</cmath> | ||
Hence, <math>PB = \frac{ab}{b+c}</math> or <math>\frac{ac}{b+c}</math>, and so the <math>P</math> corresponding to these lengths are our answer. | Hence, <math>PB = \frac{ab}{b+c}</math> or <math>\frac{ac}{b+c}</math>, and so the <math>P</math> corresponding to these lengths are our answer. | ||
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{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:29, 10 May 2023
Problem
Let be a triangle. Find all points
on segment
satisfying the following property: If
and
are the intersections of line
with the common external tangent lines of the circumcircles of triangles
and
, then
Solution
Let circle (i.e. the circumcircle of
),
be
with radii
,
and centers
, respectively, and
be the distance between their centers.
Lemma.
Proof. Let the external tangent containing meet
at
and
at
, and let the external tangent containing
meet
at
and
at
. Then clearly
and
are parallel (for they are both perpendicular
), and so
is a trapezoid.
Now, by Power of a Point, and so
is the midpoint of
. Similarly,
is the midpoint of
. Hence,
Let
,
meet
s at
, respectively. Then by similar triangles and the Pythagorean Theorem we deduce that
and
. But it is clear that
,
is the midpoint of
,
, respectively, so
as desired.
Lemma 2. Triangles and
are similar.
Proof. and similarly
, so the triangles are similar by AA Similarity.
Also, let intersect
at
. Then obviously
is the midpoint of
and
is an altitude of triangle
.Thus, we can simplify our expression of
:
where
is the length of the altitude from
in triangle
. Hence, substituting into our condition and using
gives
Using
by Heron's Formula (where
is the area of triangle
, our condition becomes
which by
becomes
Let
; then
. The quadratic in
is
which factors as
Hence,
or
, and so the
corresponding to these lengths are our answer.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.