AoPS - Problem

Problem

Source: IMO ShortList 2001, geometry problem 6

Tags: geometry, linear algebra, area, Triangle, IMO Shortlist

orl

30.09.2004 20:44

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.

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Zhero

27.02.2011 06:40

Because affine transformations preserve ratios of areas, it is sufficient to prove this when $A = (0,0)$, $B = (1,0)$, and $C = (0,1)$, and we may suppose without loss of generality that $P$ and $A$ do not lie on the same side as $BC$. Hence, we may suppose that $P = (p,-q)$, where $q$ is a positive real number. Consider the attached diagram with colors. $P$ cannot lie in the left pink region, since then $\triangle PBD$ lies inside $\triangle PAF$; $P$ cannot lie in the right pink region, since then $\triangle PCE$ lies inside $\triangle PBD$; $P$ cannot lie in the orange region, since then $\triangle PBD$ lies on $\triangle PAF$; and $P$ cannot lie in the yellow region, since then $\triangle PBD$ lies inside $\triangle PAF$, so $P$ must lie in the green region. Therefore, $p \geq 1$ and $p - q \leq 1$. We compute the coordinates of $D$, $E$, and $F$. Line $AP$ has equation $y = \frac{1+q}{p}x + 1$, which intersects the line $y=0$ at $D = \left(\frac{p}{q+1}, 0\right)$; line $CP$ has equation $y = \frac{q}{1-p}x + \frac{q}{1-p}$, and it intersects the line $x=0$ at $F = \left(0, \frac{q}{1-p}\right)$; and line $BP$ has equation $y = \frac{-q}{p}x$, and intersects the line $x+y=1$ at $E = \left(\frac{p}{p-q}, \frac{-q}{p-q}\right)$. It is easy to see that the area of $\triangle PBD$ is $\frac{pq}{2(q+1)}$ and that the area of $\triangle PFA$ is $(1/2)p\left(\frac{q}{p-1} - 1\right) = \frac{p(q-p+1)}{2}$. The distance from $P$ to the line $x+y=1$ is $\frac{1+q-p}{\sqrt{2}}$, and the length of $CE$ is $\frac{\sqrt{2} q}{p-q}$, so the area of $\triangle PCE$ is $\frac{(1+q-p)q}{2(p-q)}$. Equating these areas, we find that \begin{align*} \frac{pq}{q+1} = \frac{p(1+q-p)}{p-1} = \frac{(1+q-p)q}{p-q} = 2K, \tag{1} \end{align*} where $K$ is the area of $\triangle PBD$, $\triangle PCE$, and $\triangle PAF$. From the first and second equations, we get $\frac{q}{q+1} = \frac{q}{p-1} - 1$, so $\frac{q}{p-1} = \frac{2q+1}{q+1}$, so $p-1 = \frac{q^2 + q}{2q+1}$, so \begin{align*} p = \frac{q^2 + 3q + 1}{2q+1}. \tag{2} \end{align*} From the second and third equations, we get $\frac{p}{p-1} = \frac{q}{p-q}$, or $p^2 - 2pq + q = 0$. Substituting in the equation from (2) and multiplying both sides by $(2q+1)^2$, we find that \begin{align*} 0 &= (2q+1)^2 (p^2 - 2pq + q) \\ &= (p(2q+1))^2 - 2q(2q+1)(p(2q+1)) + q(2q+1)^2 \\ &= (q^2 + 3q + 1)^2 - (4q^2 + 2q)(q^2 + 3q + 1) + q(4q^2 + 4q + 1) \\ &= (q^4 + 6q^3 + 11q^2 + 6q + 1) - (4q^4 + 14q^3 + 10q^2 + 2q) + 4q^3 + 4q^2 + q \\ &= -3q^4 - 4q^3 + 5q^2 + 5q + 1 \\ &= -(3q+1)(q^3+q^2-2q-1). \end{align*} Because $q > 0$, $q \neq \frac{-1}{3}$, so $q^3 + q^2 - 2q - 1 = 0$. Hence, $q^3 + 3q^2 + q = 2q^2 + 3q + 1 = (q+1)(2q+1)$, so $\frac{q+1}{q} = \frac{q^2 + 3q + 1}{2q+1} = p$, by (2), so $\frac{pq}{q+1} = 1$. But from (1), we get that $K = \frac{1}{2}$, which is the area of $\triangle ABC$, so we are done.

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Mewto55555

09.07.2011 07:54

Since coordinates are so nasty and ugly, we use

Dragonboy

01.02.2012 15:35

Mewto55555 wrote: Since $p+q+r=1$, we have $\frac{qr}{1-r}=\frac{pq}{1-q}=\frac{pr}{1-p}=K$ I'm confused. Because, it is possible that some of these ratios are positive and some of them are negative. For example: it is possible: $\frac{qr}{1-r}=\frac{pq}{1-q}=1$ and $\frac{pr}{1-p}=-1$.

Mahi

01.06.2012 20:50

Dragonboy wrote: I'm confused. Because, it is possible that some of these ratios are positive and some of them are negative. For example: it is possible: $\frac{qr}{1-r}=\frac{pq}{1-q}=1$ and $\frac{pr}{1-p}=-1$. Just notice the directions, $[PBD], [PCE],[PAF]$, which are opposite to direction of $[ABC]$. So, we can safely assume that ratios of their areas will have the same sign (in this case, negative, as the direction is opposite).

dk911

28.10.2019 13:08

Is there a solution by complex coordinates? I think we can easily calculate areas by matrices

mathaddiction

20.11.2020 10:09

WLOG assume $P$ lies inside $\angle ABC$. We first deal with a few configuration issues. CLAIM 1. $P$ and $B$ lies in different sides of $AC$. Proof. Otherwise $\triangle PBD$ lies strictly inside $\triangle PAF$, hence $[PBD]<[PAF]$, contradiction. $\blacksquare$ CLAIM 2. $A$ lies between $F$ and $B$. Proof. Otherwise $\triangle PCE$ lies strictly inside $\triangle PAF$, hence $[PBD]<[PAF]$, contradiction. $\blacksquare$ Now let $A=(1:0:0)$, $B=(0:1:0)$ and $C=(0:0:1)$. Let $P=(x:y:z)$, where $x+y+z=1$, we have $$D=(0:y:z),\hspace{5pt}E=(x:0:z)\text{ and } F=(x:y:0)$$Then from the above assumptions, $x>1>z>0>y$ Therefore, \begin{align*} \frac{[PCE]}{[ABC]}&=\begin{vmatrix}x&0&\frac{x}{x+z}\\y&0&0\\z&1&\frac{z}{x+z}\end{vmatrix}\\ &=\frac{xy}{x+z} \end{align*}Notice that $\frac{xy}{x+z}<0$, hence $|[PCE]|=-[ABC]\cdot\frac{xy}{x+z}$, similarly we have $|[PBD]|=-[ABC]\cdot\frac{xz}{y+z}$ and $|[PAF]|=-[ABC]\cdot\frac{yz}{x+y}$ Therefore we have $$\frac{xy}{x+z}=\frac{xz}{y+z}=\frac{yz}{x+y}\hspace{5pt}(1)$$This becomes $x(x+y)=y(y+z)=z(x+z)$. Hence $$z=\frac{x^2+xy-y^2}{y}=x^2+x-y \hspace{5pt}(2)$$Therefore from $x+y+z=1$ we have $2x+\frac{x^2}{y}=1$ hence $$y=\frac{x^2}{1-2x}\hspace{5pt}(3)$$Substituting $(2)$ into $(1)$ we have $$z=\frac{1-3x+x^2}{1-2x}\hspace{5pt}(4)$$Now we have $x(x+y)=z(z+x)$, hence $$x\left(x+\frac{x^2}{1-2x}\right)=\frac{1-3x+x^2}{1-2x}\cdot\left(\frac{1-3x+x^2}{1-2x}+x\right)$$Clearing denominators, we have \begin{align*} 3x^4-4x^3-5x^2+5x-1&=0\\ (3x-1)(x^3-x^2-2x+1)&=0\\ x^3-x^2-2x+1&=0 \end{align*}Therefore, $$\frac{xz}{y+z}=\frac{x(1-3x+x^2)}{(1-x)(1-2x)}=-1$$This completes the proof.

Phorphyrion

25.10.2022 21:40

A more symmetric way to solve the equations after using bary:

awesomeming327.

16.10.2023 20:10

We use barycentric coordinates. Let $P$ be $(\tfrac{x}{x+y+z},\tfrac{y}{x+y+z},\tfrac{z}{x+y+z})$. We note that $\triangle BDP$, $\triangle AFP$, and $\triangle CEP$ are all oppositely oriented to $\triangle ABC$, so where $[s]$ denotes signed area, we'll show that \[\frac{[AFP]}{[ABC]}=\frac{[BDP]}{[ABC]}=\frac{[CEP]}{[ABC]}=t\]implies $t=-1$. Note that $PC$ intersects $AB$ at $F=(\tfrac{x}{x+y},\tfrac{y}{x+y},0)$, and $PA$ intersects $BC$ at $D=(0,\tfrac{y}{y+z},\tfrac{z}{y+z})$, and $PB$ intersects $CA$ at $E=(\frac{x}{z+x},0,\frac{z}{z+x})$. We have \[\frac{[AFP]}{[ABC]} = \begin{vmatrix} 1 & 0 & 0\\ \frac{x}{x+y} & \frac{y}{x+y} & 0\\ \frac{x}{x+y+z} & \frac{y}{x+y+z} & \frac{z}{x+y+z} \end{vmatrix} = \frac{yz}{(x+y+z)(x+y)} \]\[ \frac{[BDP]}{[ABC]} = \begin{vmatrix} 0 & 1 & 0\\ 0 & \frac{y}{y+z} & \frac{z}{y+z}\\ \frac{x}{x+y+z} & \frac{y}{x+y+z} & \frac{z}{x+y+z} \end{vmatrix} = \frac{zx}{(x+y+z)(y+z)} \]\[ \frac{[CEP]}{[ABC]} = \begin{vmatrix} 0 & 0 & 1\\ \frac{x}{z+x} & 0 & \frac{z}{z+x}\\ \frac{x}{x+y+z} & \frac{y}{x+y+z} & \frac{z}{x+y+z} \end{vmatrix} = \frac{xy}{(x+y+z)(z+x)}\]Since everything is homogeneous, we can assume $x+y+z=1$. Now, we have \[\frac{xyz}{x(x+y)}=\frac{xyz}{y(y+z)}=\frac{xyz}{z(z+x)}\]so $x(x+y)=y(1-x)=(1-x-y)(1-y)$. Expanding, we have $x^2+xy=y-yx=1-x-2y+xy+y^2$. We have $x^2=y(1-2x)\implies y=\frac{x^2}{1-2x}$ because obviously $x\neq \tfrac12$ and we have $x^2+x-1=y^2-2y=\frac{x^2(x^2+4x-2)}{(1-2x)^2}$. This implies that $(3x-1)(x^3-x^2-2x+1)=3x^4-4x^3-5x^2+5x-1=0$. $x=\tfrac13$ implies $y=\tfrac13$ and $z=\tfrac13$ but at least one of $x$, $y$, $z$ must be negative. Thus, \[x^2\left(x-1+\frac{1}{y}\right)=x^3-x^2-2x+1=0\]Clearly $x$ cannot be zero, so $1-x=\tfrac1y$. Thus, $y(y+z)=y(1-x)=1$ so we have \[\frac{zx}{(x+y+z)(y+z)}=xyz\]so it remains to show that $xyz=-1$. The cyclic equivalents of $1-x=\tfrac1y$ are $1-y=\tfrac1z$ and $1-z=\tfrac1x$. If $x=y$ then we get $x^2-x-1=0$ and it is easy to see that this is not a solution to $x^3-x^2-2x+1=0$. Thus, $x$, $y$, $z$ are mutually distinct. Thus, being the three roots of $f(x)=x^3-x^2-2x+1$, by Vieta's formulas we must have $xyz=-1$. We're done.