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Difference between revisions of "2001 IMO Shortlist Problems/G6"
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== Solution ==
 
== Solution ==
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{{solution}}
 
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[hide="Solution 1 by Mewto55555"]
 
  
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We use
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Solution 1 by Mewto55555:
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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>.
 
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>.
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Thus, <math>K=\frac{pq}{1-q}=\frac{\frac{p}{1-p}}{1-\frac{1}{1-p}}=\frac{p}{1-p-1}=-1</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>
  
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Therefore, if <math>[PBD]=[PCE]=[PAF]</math>, necessarily <math>[PBD]=[PCE]=[PAF]=[ABC]</math>.[/hide]
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Therefore, if <math>[PBD]=[PCE]=[PAF]</math>, necessarily <math>[PBD]=[PCE]=[PAF]=[ABC]</math>.
  
 
== Resources ==
 
== Resources ==

Latest revision as of 23:25, 2 April 2012

Problem

Let $ABC$ be a triangle and $P$ an exterior point in the plane of the triangle. Suppose the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ (or extensions thereof) in $D$, $E$, $F$, respectively. Suppose further that the areas of triangles $PBD$, $PCE$, $PAF$ are all equal. Prove that each of these areas is equal to the area of triangle $ABC$ itself.

Solution

Solution 1 by Mewto55555:

We use barycentric coordinates.

So $A$ is $(1,0,0)$, $B$ is $(0,1,0)$, $C$ is $(0,0,1)$, and $P$ is $(p,q,r)$, with $p+q+r=1$.

Now, the equation of line $AP$ is just the line $qz=ry$, $BP$ is just $pz=rx$, and $CP$ is $qx=py$.

Also, $AB$ is just $z=0$, $BC$ is $x=0$, and $AC$ is $y=0$.

Thus, the coordinates of $D$ is $\left(0,\frac{q}{q+r},\frac{r}{q+r}\right)$. Similarly, $E$ is at $\left(\frac{p}{p+r},0,\frac{r}{p+r}\right)$ and $F$ is at $\left(0,\frac{q}{q+r},\frac{r}{q+r}\right)$

Now, the ratio $[PBD]$ to $[ABC]$ is just

$\begin{vmatrix} p & 0 & 0 \\ q & 1 & \frac{q}{q+r}\\ r & 0 & \frac{r}{q+r} \end{vmatrix}= \frac{pr}{q+r}$

The other ratios are similarly $\frac{pq}{p+r}$ and $\frac{qr}{p+q}$

Since $p+q+r=1$, we have $\frac{qr}{1-r}=\frac{pq}{1-q}=\frac{pr}{1-p}=K$ and we want to show that $|K|=1$.

Thus, we have $\frac{pqr}{p(1-r)}=\frac{pqr}{r(1-q)}=\frac{pqr}{q(1-p)}$.

Since none of $p,q,r=0$ (else $P$ would be on one of the sides of $ABC$):

$p(1-r)=r(1-q)=q(1-p)$.

We know $r=1-p-q$. Substuting:

$p^2+pq=1-p-2q+pq+q^2=q-pq$.

From the first and third, we get that $q(1-2p)=p^2 \implies q=\frac{p^2}{1-2p}$

Now consider first and second;

$p^2+p-1=q^2-2q$

Subbing back in $q$:

$(p^2+p-1)(1-2p)^2=p^4-2p^2(1-2p)$

which rearranges to

$0=3p^4-4p^3-5p^2+5p-1=(3p-1)(p^3-p^2-2p+1)=0$

If $p=\frac{1}{3}$, then $q=r=\frac{1}{3}$, so $P$ is in the triangle (as all of $p,q,r>0$) contradiction.

Thus, we have $p^3-p^2-2p+1=0$

So, $1-2p=p^2(1-p) \implies q=\frac{p^2}{1-2p}=\frac{1}{1-p}$

Thus, $K=\frac{pq}{1-q}=\frac{\frac{p}{1-p}}{1-\frac{1}{1-p}}=\frac{p}{1-p-1}=-1$

Therefore, if $[PBD]=[PCE]=[PAF]$, necessarily $[PBD]=[PCE]=[PAF]=[ABC]$.

Resources

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