Difference between revisions of "2020 USAMO Problems/Problem 1"
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==Problem 1== | ==Problem 1== | ||
Let <math>ABC</math> be a fixed acute triangle inscribed in a circle <math>\omega</math> with center <math>O</math>. A variable point <math>X</math> is chosen on minor arc <math>AB</math> of <math>\omega</math>, and segments <math>CX</math> and <math>AB</math> meet at <math>D</math>. Denote by <math>O_1</math> and <math>O_2</math> the circumcenters of triangles <math>ADX</math> and <math>BDX</math>, respectively. Determine all points <math>X</math> for which the area of triangle <math>OO_1O_2</math> is minimized. | Let <math>ABC</math> be a fixed acute triangle inscribed in a circle <math>\omega</math> with center <math>O</math>. A variable point <math>X</math> is chosen on minor arc <math>AB</math> of <math>\omega</math>, and segments <math>CX</math> and <math>AB</math> meet at <math>D</math>. Denote by <math>O_1</math> and <math>O_2</math> the circumcenters of triangles <math>ADX</math> and <math>BDX</math>, respectively. Determine all points <math>X</math> for which the area of triangle <math>OO_1O_2</math> is minimized. | ||
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+ | ==Solution== | ||
+ | [[File:2020 USAMO 1.png|400px|right]] | ||
+ | Let <math>E</math> be midpoint <math>AD.</math> Let <math>F</math> be midpoint <math>BD \implies</math> | ||
+ | <cmath>EF = ED + FD = \frac {AD}{2} + \frac {BD}{2} = \frac {AB}{2}.</cmath> | ||
+ | <math>E</math> and <math>F</math> are the bases of perpendiculars dropped from <math>O_1</math> and <math>O_2,</math> respectively. | ||
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+ | Therefore <math>O_1O_2 \ge EF = \frac {AB}{2}.</math> | ||
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+ | <cmath>CX \perp O_1O_2, AX \perp O_1O \implies \angle O O_1O_2 = \angle AXC</cmath> | ||
+ | <math>\angle AXC = \angle ABC (AXBC</math> is cyclic) <math>\implies \angle O O_1O_2 = \angle ABC.</math> | ||
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+ | Similarly <math>\angle BAC = \angle O O_2 O_1 \implies \triangle ABC \sim \triangle O_2 O_1O.</math> | ||
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+ | The area of <math>\triangle OO_1O_2</math> is minimized if <math>CX \perp AB</math> because | ||
+ | <math></math>\frac {[OO_1O_2]} {[ABC]} = (\frac {O_1 O_2} {AB})^2 \ge (\frac {EF} {AB})^2 = \frac {1}{4}$. | ||
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Revision as of 15:29, 15 September 2022
Problem 1
Let be a fixed acute triangle inscribed in a circle
with center
. A variable point
is chosen on minor arc
of
, and segments
and
meet at
. Denote by
and
the circumcenters of triangles
and
, respectively. Determine all points
for which the area of triangle
is minimized.
Solution
Let be midpoint
Let
be midpoint
and
are the bases of perpendiculars dropped from
and
respectively.
Therefore
is cyclic)
Similarly
The area of is minimized if
because
$$ (Error compiling LaTeX. Unknown error_msg)\frac {[OO_1O_2]} {[ABC]} = (\frac {O_1 O_2} {AB})^2 \ge (\frac {EF} {AB})^2 = \frac {1}{4}$.