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

Given $\triangle ABC$. On the side AB externally is constructed square ABPQ. On the side AC internally is constructed square ACMN. AH is the altitude. If $O_1$ and$O_2$ are the centers of the two squares, prove that $O_1, O_2 $ and $H$ are collinear.

share|cite|improve this question
Is $H$ the intersection of the altitudes or the intersection of the altitude with the opposite side? – Phira Jun 4 '12 at 20:23
From the question, $AH$ is an altitude, so $H$ lies on $BC$. – Théophile Jun 4 '12 at 20:44
up vote 1 down vote accepted

Let $A'$ be the intersection of the altitude through $A$ with $BC$ (the point called $H$ by the OP).

Let $X$ be the point on $BC$ such that $AA'X$ forms an isosceles rectangular triangle. (The vector $BC$ points in the same direction as the vector $A'X$.)

Then, a clockwise rotation by $45^\circ$ around $A$ followed by a scaling by $\frac1 {\sqrt 2}$ moves the points $X,C,B$ to the points $A',O_2,O_1$.

Clearly, a line remains a line after rotating and scaling.

share|cite|improve this answer
Nicely done. A couple of comments, though: first, why rename (and redefine) $H$? More importantly, there are two possibilities for $X$, one on either side of $H$. You can avoid this problem by defining $X$ to lie on $CH$. – Théophile Jun 4 '12 at 21:00
I redefined it because $H$ always denotes the orthocenter. It is like starting a question with "Let $i$ denote the square root of $-2$.". As to your other remark, I considered defining $X$ by a rotation, but it is a convention that the points of a triangle are usually given in counter-clockwise order which fixes the position of $X$. I don't understand your suggestion, because $X$ does not necessarily lie on the segment $CH$ or on the half-line $CH$. And the line $CH$ is no different from the line $BC$ that I used. But I will add a clarification. – Phira Jun 4 '12 at 21:28
Beautiful solution indeed. Do you have any idea how can I avoid using the rotation? – Adam Jun 4 '12 at 23:29

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.