Difference between revisions of "2017 USAJMO Problems/Problem 3"
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==Solution 2== | ==Solution 2== | ||
+ | We will use barycentric coordinates and vectors. Let <math>\vec{X}</math> be the position vector of a point <math>X.</math> The point <math>(x, y, z)</math> in barycentric coordinates denotes the point <math>x\vec{A} + y\vec{B} + z\vec{C}.</math> For all points in the plane of <math>\triangle{ABC},</math> we have <math>x + y + z = 1.</math> | ||
+ | |||
+ | The following are obvious: | ||
<cmath>A = (1, 0, 0)</cmath> | <cmath>A = (1, 0, 0)</cmath> | ||
<cmath>B = (0, 1, 0)</cmath> | <cmath>B = (0, 1, 0)</cmath> | ||
− | <cmath>C = (0, 0, 1)</cmath> | + | <cmath>C = (0, 0, 1).</cmath> |
− | <cmath>P = \left(x_P, y_P, z_P\right)</cmath> | + | Define the point <math>P</math> as follows: |
− | The fact that <math>P</math> lies on the circumcircle of <math>\triangle{ABC}</math> gives us <math>x^2_P + y^2_P + z^2_P = 1.</math> This, along with <math>x_P + y_P + z_P = 1 | + | <cmath>P = \left(x_P, y_P, z_P\right).</cmath> |
+ | The fact that <math>P</math> lies on the circumcircle of <math>\triangle{ABC}</math> gives us <math>x^2_P + y^2_P + z^2_P = 1.</math> This, along with the condition <math>x_P + y_P + z_P = 1</math> inherent to barycentric coordinates, gives us <math>x_Py_P + y_Pz_P + z_Px_P = 0.</math> | ||
<cmath>D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)</cmath> | <cmath>D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)</cmath> |
Revision as of 22:09, 5 May 2017
Contents
[hide]Problem
() Let
be an equilateral triangle and let
be a point on its circumcircle. Let lines
and
intersect at
; let lines
and
intersect at
; and let lines
and
intersect at
. Prove that the area of triangle
is twice that of triangle
.
Solution 1
WLOG, let . Let
, and
. After some angle chasing, we find that
and
. Therefore,
~
.
Lemma 1: If , then
.
This lemma results directly from the fact that
~
;
, or
.
Lemma 2: .
We see that
, as desired.
Lemma 3: .
We see that
However, after some angle chasing and by the Law of Sines in
, we have
, or
, which implies the result.
By the area lemma, we have and
.
We see that . Thus, it suffices to show that
, or
. Rearranging, we find this to be equivalent to
, which is Lemma 3, so the result has been proven.
Solution 2
We will use barycentric coordinates and vectors. Let be the position vector of a point
The point
in barycentric coordinates denotes the point
For all points in the plane of
we have
The following are obvious:
Define the point
as follows:
The fact that
lies on the circumcircle of
gives us
This, along with the condition
inherent to barycentric coordinates, gives us
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2017 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |