Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
This is not true. – dtldarek Jun 24 '12 at 9:13

Disproof without words:

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


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 $(9)$.

share|cite|improve this answer
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.

share|cite|improve this answer
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.

share|cite|improve this answer
P, Q and R are on the sides of the triangle. – Adam Jun 24 '12 at 18:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.