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I need a non-vector solution to the following problem: Given a triangle ABC. P, Q and R are points on the sides AB, BC and CA respectively and are such that $AP:PB=BQ:QC=CR:RA$. If $\triangle PQR$ is isosceles show that $\triangle ABC$ is also isosceles.

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This is not true. –  dtldarek Jun 24 '12 at 9:13

3 Answers 3

Disproof without words:

$\hspace{2.4cm}$enter image description here

However...

Suppose the ratios $$ \frac{|AP|}{|PB|}=\frac{|BQ|}{|QC|}=\frac{|CR|}{|RA|}=\alpha\tag{1} $$ where $\alpha\in\{\frac12,1,2\}$. If $\triangle PQR$ is isosceles, then $\triangle ABC$ is also isosceles.

Note that $$ \frac{1}{\alpha+1}\begin{bmatrix}1&\alpha&0\\0&1&\alpha\\\alpha&0&1\end{bmatrix}\begin{bmatrix}A\\B\\C\end{bmatrix}=\begin{bmatrix}P\\Q\\R\end{bmatrix}\tag{2} $$ So that $$ \begin{bmatrix}A\\B\\C\end{bmatrix}=\frac{\alpha+1}{\alpha^3+1}\begin{bmatrix}1&-\alpha&\alpha^2\\\alpha^2&1&-\alpha\\-\alpha&\alpha^2&1\end{bmatrix}\begin{bmatrix}P\\Q\\R\end{bmatrix}\tag{3} $$ The matrices in $(2)$ and $(3)$ commute with affine transformations (linear transformations + translations) because the sum of the row elements is $1$. Suppose that $M$ is a $2\times2$ matrix and $T$ is an offset (a point). Then, by the associativity and distributivity of matrix multiplication, $$ \begin{align} &\frac{\alpha+1}{\alpha^3+1}\begin{bmatrix}1&-\alpha&\alpha^2\\\alpha^2&1&-\alpha\\-\alpha&\alpha^2&1\end{bmatrix}\left(\begin{bmatrix}P\\Q\\R\end{bmatrix}M+\begin{bmatrix}T\\T\\T\end{bmatrix}\right)\\ &=\left(\frac{\alpha+1}{\alpha^3+1}\begin{bmatrix}1&-\alpha&\alpha^2\\\alpha^2&1&-\alpha\\-\alpha&\alpha^2&1\end{bmatrix}\begin{bmatrix}P\\Q\\R\end{bmatrix}\right)M+\begin{bmatrix}T\\T\\T\end{bmatrix}\\ &=\begin{bmatrix}A\\B\\C\end{bmatrix}M+\begin{bmatrix}T\\T\\T\end{bmatrix}\tag{4} \end{align} $$ Thus, applying an affine transformation to $\triangle ABC$ applies that same transformation to $\triangle PQR$, and vice versa.

Up to rotation, scaling, and translation, every isosceles triangle looks like $$ \begin{bmatrix}P\\Q\\R\end{bmatrix}=\begin{bmatrix}0&0\\1&x\\2&0\end{bmatrix}\tag{5} $$ Applying $(3)$ to $(5)$ gives $$ \begin{align} \begin{bmatrix}A\\B\\C\end{bmatrix} &=\frac{\alpha+1}{\alpha^3+1}\begin{bmatrix}1&-\alpha&\alpha^2\\\alpha^2&1&-\alpha\\-\alpha&\alpha^2&1\end{bmatrix}\begin{bmatrix}P\\Q\\R\end{bmatrix}\\ &=\frac{\alpha+1}{\alpha^3+1}\begin{bmatrix}1&-\alpha&\alpha^2\\\alpha^2&1&-\alpha\\-\alpha&\alpha^2&1\end{bmatrix}\begin{bmatrix}0&0\\1&x\\2&0\end{bmatrix}\\ &=\frac{\alpha+1}{\alpha^3+1}\begin{bmatrix}2\alpha^2-\alpha&-\alpha x\\1-2\alpha&x\\\alpha^2+2&\alpha^2x\end{bmatrix}\tag{6} \end{align} $$ Plugging $\alpha\in\{\frac12,1,2\}$ into $(6)$ yields $$ \begin{bmatrix}A\\B\\C\end{bmatrix}=\frac13\begin{bmatrix}0&-2x\\0&4x\\5&x\end{bmatrix}\quad\text{for }\alpha=\frac12\tag{7} $$ $$ \begin{bmatrix}A\\B\\C\end{bmatrix}=\begin{bmatrix}1&-x\\-1&x\\3&x\end{bmatrix}\quad\text{for }\alpha=1\tag{8} $$ $$ \begin{bmatrix}A\\B\\C\end{bmatrix}=\frac13\begin{bmatrix}6&-2x\\-3&x\\6&4x\end{bmatrix}\quad\text{for }\alpha=2\tag{9} $$ Note that $|AC|=|BC|$ in $(7)$, $|AB|=|AC|$ in $(8)$, and $|AB|=|BC|$ in $(9)$.

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Without labels, too, but I take it the outside triangle is ABC, the inside, PQR. Well, ABC looks isosceles to me, so how is this a disproof of anything? –  Gerry Myerson Jun 24 '12 at 23:59
    
@GerryMyerson: Sorry about that. I had thought the grid would be enough. –  robjohn Jun 25 '12 at 11:18
    
I repeat: ABC looks isosceles to me, so how is this a disproof of anything? –  Gerry Myerson Jun 25 '12 at 12:31
    
@GerryMyerson: Oh! I was reading the implication backwards! I have to go out for a while, but when I get back, I will put the image above through the proper linear map to make $\triangle PQR$ isosceles. –  robjohn Jun 25 '12 at 13:31
    
@GerryMyerson: Sorry about that! This should be better. –  robjohn Jun 25 '12 at 15:36

The statement is wrong. Consider the triangle $\Delta$ with vertices $$A_0=(-16,-8),\quad A_1=(20, -4),\quad A_2=(-4,12)\ .$$ Its center of gravity is at $O=(0,0)$, and the three lengths $OA_i$ are different, so $\Delta$ is not isosceles.

Now put $$B_i:={1\over4}A_{i-1}+{3\over4}A_{i+1}\qquad(i=0,1,2)\ .$$ This gives $$B_0=(14,0),\quad B_1=(-7,7),\quad B_2=(-7,-7)\ ,$$ and the triangle $\Delta'$ with vertices $B_0$, $B_1$, $B_2$ is isosceles.

A copy of triangle $\Delta$ is pictured in robjohn's answer.

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I had not noticed that ours were the same! I made mine with the isosceles triangle pointing up. Noting that $$ \begin{bmatrix}\frac14&\frac34&0\\ 0&\frac14&\frac34\\ \frac34&0&\frac14\end{bmatrix} \begin{bmatrix}A\\B\\C\end{bmatrix} =\begin{bmatrix}P\\Q\\R\end{bmatrix} $$ I computed $$ \begin{bmatrix}A\\B\\C\end{bmatrix} =\begin{bmatrix}\frac14&\frac34&0\\ 0&\frac14&\frac34\\ \frac34&0&\frac14\end{bmatrix}^{-1} \begin{bmatrix}P\\Q\\R\end{bmatrix} =\begin{bmatrix}\frac17&-\frac37&\frac97\\ \frac97&\frac17&-\frac37\\ -\frac37&\frac97&\frac17\end{bmatrix} \begin{bmatrix}P\\Q\\R\end{bmatrix} $$ –  robjohn Jun 25 '12 at 19:12
    
So I made the coordinates of $\triangle PQR$ divisible by $7$: $(0,0)$, $(7,21)$, and $(14,0)$; and applied the matrix above to get $\triangle ABC$. Since horizontally oriented graphics fit better in the answers, I rotated everything 90°. Nice answer :-) (+1) –  robjohn Jun 25 '12 at 19:12

Maybe I'm missing something, but the statement looks false. For example:

$A = (0, 0)$, $B = (4, 0)$, $C = (0, 6)$, $P = (2, 1)$, $Q = (2, 3)$, $R = (0, 3)$

Then $AP = PB$, $BQ = QC$, and $CR = RA$. Also $\triangle PQR$ is isosceles, but $\triangle ABC$ is not.

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P, Q and R are on the sides of the triangle. –  Adam Jun 24 '12 at 18:11

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