Difference between revisions of "2017 USAJMO Problems/Problem 3"
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Extend <math>DP</math> to hit <math>EF</math> at <math>K</math>. Then note that <math>[DEF]\cdot\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=[ABC].</math> Letting <math>BF=x</math> and <math>PF=y</math>, we have that <math>\frac{x+AB}y=\frac{y+PC}x=\frac{AC}{BP}.</math> Solving and simplifying using LoC on <math>\triangle BPC</math> gives <math>\frac{AB}{AF}=\frac{PC}{PB+PC}.</math> Similarly, <math>\frac{AC}{AE}=\frac{PB}{PB+PC}.</math> | Extend <math>DP</math> to hit <math>EF</math> at <math>K</math>. Then note that <math>[DEF]\cdot\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=[ABC].</math> Letting <math>BF=x</math> and <math>PF=y</math>, we have that <math>\frac{x+AB}y=\frac{y+PC}x=\frac{AC}{BP}.</math> Solving and simplifying using LoC on <math>\triangle BPC</math> gives <math>\frac{AB}{AF}=\frac{PC}{PB+PC}.</math> Similarly, <math>\frac{AC}{AE}=\frac{PB}{PB+PC}.</math> | ||
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− | Now we find <math>\frac{AK}{DK}.</math> Note that <math>\frac{AD}{DP}=\frac{AD}{BD}\cdot\frac{BD}{DP}=\frac{AC}{PB}\cdot\frac{AB}{PC}=\frac{AB^2}{PB\cdot PC}.</math> | + | Now we find <math>\frac{AK}{DK}.</math> Note that <math>\frac{AD}{DP}=\frac{AD}{BD}\cdot\frac{BD}{DP}=\frac{AC}{PB}\cdot\frac{AB}{PC}=\frac{AB^2}{PB\cdot PC}.</math> Now let <math>E'=DE\cap AF</math> and <math>F'=DF\cap AE</math>. Then by an area/concurrence theorem, we have that <math>\frac{DK}{AK}+\frac{DE'}{EE'}+\frac{DF'}{FF'}=1,</math> or <math>\frac{DK}{AK}+(1-\frac{DP}{AP}-\frac{DC}{BC})+(1-\frac{DP}{AP}-\frac{BD}{BC})=1.</math> Thus we have that <math>\frac{DK}{AK}=2\cdot\frac{DP}{AP}.</math> |
− | + | Manipulating these gives <math>\frac{AK}{DK}=\frac{(PB+PC)^2}{2\cdot PB\cdot PC}.</math> Thus <math>\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=\frac12,</math> and we are done. | |
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− | Manipulating these gives <math>\frac{ | ||
~cocohearts | ~cocohearts | ||
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By Mathdummy. | By Mathdummy. | ||
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==See also== | ==See also== | ||
{{USAJMO newbox|year=2017|num-b=2|num-a=4}} | {{USAJMO newbox|year=2017|num-b=2|num-a=4}} |
Latest revision as of 03:13, 16 March 2022
Contents
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 (No Bash)
Extend to hit
at
. Then note that
Letting
and
, we have that
Solving and simplifying using LoC on
gives
Similarly,
Now we find Note that
Now let
and
. Then by an area/concurrence theorem, we have that
or
Thus we have that
Manipulating these gives Thus
and we are done.
~cocohearts
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
It is clear that
;
; and
Define the point as
The fact that
lies on the circumcircle of
gives us
This, along with the condition
inherent to barycentric coordinates, gives us
We can write the equations of the following lines:
We can then solve for the points :
The area of an arbitrary triangle is:
To calculate we wish to compute
After a lot of computation, we obtain the following:
Evaluating the denominator,
Since and
it follows that:
We thus conclude that:
From this, it follows that and we are done.
Solution 3
![[asy] import cse5; import graph; import olympiad; size(3inch); pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3)); pair P = (-1, -sqrt(11)+sqrt(3)); path circle = Circle(O, 2sqrt(3)); pair D = extension(A,P,B,C); pair E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label("A", A, N); label("B", B, W); label("C", C, E); label("P", P, S); label("D", D, NW); label("E", E1, SE); label("F", F, SW); dot("O", O, SE); [/asy]](http://latex.artofproblemsolving.com/1/0/d/10d76c13207b6356c0ebc7833813448e1e89cd04.png)
We'll use coordinates and shoelace. Let the origin be the midpoint of . Let
, and
, then
. Using the facts
and
, we have
, so
, and
.
The slope of is
It is well-known that
is self-polar, so
is the polar of
, i.e.,
is perpendicular to
. Therefore, the slope of
is
. Since
, we get the x-coordinate of
,
, i.e.,
. Using shoelace,
So
. Q.E.D
By Mathdummy.
Solution 4 Without the nasty computations
Note that . We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths.
Let and
. Then,
From Law of Cosine,
.
From Ptolemy's theorem, , so
.
Lemma 1: In Triangle ABC with side lengths and
, the length of the angle bisector of
is
This can be easily proved with Stewart's and Law of Cosine.
Using Lemma 1, we have
Plug in
, we get:
Then
By Mathdummy.
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 |