Difference between revisions of "2001 IMO Shortlist Problems/G6"
(New page: == Problem == Let <math>ABC</math> be a triangle and <math>P</math> an exterior point in the plane of the triangle. Suppose the lines <math>AP</math>, <math>BP</math>, <math>CP</math> meet...) |
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== Solution == | == Solution == | ||
− | {{ | + | |
+ | Solution 1 by Mewto55555: | ||
+ | |||
+ | We use barycentric coordinates. | ||
+ | |||
+ | So <math>A</math> is <math>(1,0,0)</math>, <math>B</math> is <math>(0,1,0)</math>, <math>C</math> is <math>(0,0,1)</math>, and <math>P</math> is <math>(p,q,r)</math>, with <math>p+q+r=1</math>. | ||
+ | |||
+ | Now, the equation of line <math>AP</math> is just the line <math>qz=ry</math>, <math>BP</math> is just <math>pz=rx</math>, and <math>CP</math> is <math>qx=py</math>. | ||
+ | |||
+ | Also, <math>AB</math> is just <math>z=0</math>, <math>BC</math> is <math>x=0</math>, and <math>AC</math> is <math>y=0</math>. | ||
+ | |||
+ | Thus, the coordinates of <math>D</math> is <math>\left(0,\frac{q}{q+r},\frac{r}{q+r}\right)</math>. Similarly, <math>E</math> is at <math>\left(\frac{p}{p+r},0,\frac{r}{p+r}\right)</math> and <math>F</math> is at <math>\left(0,\frac{q}{q+r},\frac{r}{q+r}\right)</math> | ||
+ | |||
+ | Now, the ratio <math>[PBD]</math> to <math>[ABC]</math> is just | ||
+ | |||
+ | <math>\begin{vmatrix} | ||
+ | p & 0 & 0 \ | ||
+ | q & 1 & \frac{q}{q+r}\ | ||
+ | r & 0 & \frac{r}{q+r} | ||
+ | \end{vmatrix}= \frac{pr}{q+r}</math> | ||
+ | |||
+ | The other ratios are similarly <math>\frac{pq}{p+r}</math> and <math>\frac{qr}{p+q}</math> | ||
+ | |||
+ | Since <math>p+q+r=1</math>, we have <math>\frac{qr}{1-r}=\frac{pq}{1-q}=\frac{pr}{1-p}=K</math> and we want to show that <math>|K|=1</math>. | ||
+ | |||
+ | Thus, we have <math>\frac{pqr}{p(1-r)}=\frac{pqr}{r(1-q)}=\frac{pqr}{q(1-p)}</math>. | ||
+ | |||
+ | Since none of <math>p,q,r=0</math> (else <math>P</math> would be on one of the sides of <math>ABC</math>): | ||
+ | |||
+ | <math>p(1-r)=r(1-q)=q(1-p)</math>. | ||
+ | |||
+ | We know <math>r=1-p-q</math>. Substuting: | ||
+ | |||
+ | <math>p^2+pq=1-p-2q+pq+q^2=q-pq</math>. | ||
+ | |||
+ | From the first and third, we get that <math>q(1-2p)=p^2 \implies q=\frac{p^2}{1-2p}</math> | ||
+ | |||
+ | Now consider first and second; | ||
+ | |||
+ | <math>p^2+p-1=q^2-2q</math> | ||
+ | |||
+ | Subbing back in <math>q</math>: | ||
+ | |||
+ | <math>(p^2+p-1)(1-2p)^2=p^4-2p^2(1-2p)</math> | ||
+ | |||
+ | which rearranges to | ||
+ | |||
+ | <math>0=3p^4-4p^3-5p^2+5p-1=(3p-1)(p^3-p^2-2p+1)=0</math> | ||
+ | |||
+ | If <math>p=\frac{1}{3}</math>, then <math>q=r=\frac{1}{3}</math>, so <math>P</math> is in the triangle (as all of <math>p,q,r>0</math>) contradiction. | ||
+ | |||
+ | Thus, we have <math>p^3-p^2-2p+1=0</math> | ||
+ | |||
+ | So, <math>1-2p=p^2(1-p) \implies q=\frac{p^2}{1-2p}=\frac{1}{1-p}</math> | ||
+ | |||
+ | Thus, <math>K=\frac{pq}{1-q}=\frac{\frac{p}{1-p}}{1-\frac{1}{1-p}}=\frac{p}{1-p-1}=-1</math> | ||
+ | |||
+ | Therefore, if <math>[PBD]=[PCE]=[PAF]</math>, necessarily <math>[PBD]=[PCE]=[PAF]=[ABC]</math>. | ||
== Resources == | == Resources == |
Latest revision as of 22:25, 2 April 2012
Problem
Let be a triangle and an exterior point in the plane of the triangle. Suppose the lines , , meet the sides , , (or extensions thereof) in , , , respectively. Suppose further that the areas of triangles , , are all equal. Prove that each of these areas is equal to the area of triangle itself.
Solution
Solution 1 by Mewto55555:
We use barycentric coordinates.
So is , is , is , and is , with .
Now, the equation of line is just the line , is just , and is .
Also, is just , is , and is .
Thus, the coordinates of is . Similarly, is at and is at
Now, the ratio to is just
The other ratios are similarly and
Since , we have and we want to show that .
Thus, we have .
Since none of (else would be on one of the sides of ):
.
We know . Substuting:
.
From the first and third, we get that
Now consider first and second;
Subbing back in :
which rearranges to
If , then , so is in the triangle (as all of ) contradiction.
Thus, we have
So,
Thus,
Therefore, if , necessarily .