Let $O$ and $H$ be respectively the circumcenter and the orthocenter of triangle $ABC$. Let $a$, $b$ and $c$ denote the side lengths. We are given that $a^2+b^2+c^2=29$ and the circumradius is $R=9$. We need to find $OH^2$.

I know that there is formula $OH^2=9R^2-(a^2+b^2+c^2)$, but I cannot use it unless I prove it.

I tried placing ABC triangle in a Cartesian plane and found coordinates of $O$ and $H$, but the expressions are not nice and I didn't manage to simplify them to required result.

Maybe vectors can be used? Please, note that I cannot use sophisticated vector knowledge in the solution. Any suggestions or ideas would be appreciated.

  • $\begingroup$ why don't you prove your formula $$OH^2=9R^2-(a^2+b^2+c^2)$$? $\endgroup$ Commented Jun 21, 2015 at 18:40
  • $\begingroup$ see herehttp://www.cut-the-knot.org/arithmetic/algebra/DistanceOH.shtml#solution $\endgroup$ Commented Jun 21, 2015 at 18:42
  • $\begingroup$ @Dr. Sonnhard Graubner Thank you. $\endgroup$
    – Ekushkebi
    Commented Jun 21, 2015 at 19:03

2 Answers 2


Take the circumcenter as the origin. Then vertices $A, B ,C$ become $\vec A,\vec B$ and $\vec C$ respectively, where $|\vec A|=|\vec B|=|\vec C|$. Also, let the circumcenter be $O$, centroid $G$, and orthocenter $H$.

Now, the centroid $G$ can be written as: $$G=\dfrac {\vec A+\vec B+\vec C}3\,.$$ Also we know that $\dfrac {HG}{GO}=\dfrac 21$, which means $$HO=3\cdot GO\implies HO=3\cdot\left( \dfrac {\vec A+\vec B+\vec C}3\right)={\vec A+\vec B+\vec C}\,.$$ Therefore, $$OH^2=(\vec A+\vec B+\vec C)\cdot(\vec A+\vec B+\vec C)=3R^2+2(\vec A\cdot\vec B+\vec B\cdot\vec C+\vec C\cdot\vec A)$$ Note that $$\begin{align}\vec A\cdot\vec B+\vec B\cdot\vec C+\vec C\cdot\vec A&=R^2(\cos {2A}+\cos {2B}+\cos {2C})\\&=3R^2-2R^2(\sin^2 A+\sin^2 B+\sin^2 C)\,.\end{align}$$ Finally, using the sine rule, we get $$OH^2=9R^2-(a^2+b^2+c^2)\,.$$

The result used above is the $\textit {Euler Line Theorem}$.

$\textbf {Proof:}$

Let $O$ the circumcenter of $\triangle ABC$ and $G$ its centroid. Extend $OG$ until a point $P$ such that $OG/GP=1/2$. We'll prove that $P$ is the orthocenter $H$.

Draw the median $AA'$ where $A'$ is the midpoint of $BC$. Triangles $OGA'$ and $PGA$ are similar, since $GP=2GO$, $AG=2A'G$ and $\angle OGA'=\angle PGA$. Then $\angle OA'G =\angle PGA$ and $OA'\parallel AP$. But $OA'\perp BC$ so $AP\perp BC$, that is, $AP$ is a height of the triangle.

Repeating the same argument for the other medians proves that $P$ lies on the three heights and therefore it must be the orthocenter $H$.

The ratio is $OG/GH=1/2$ since we constructed it that way.

  • $\begingroup$ How do we know the ratio $HG/GC=2/1$, this is completely new to me :) What would be the way to prove it? $\endgroup$
    – Ekushkebi
    Commented Jun 21, 2015 at 18:54
  • 1
    $\begingroup$ Also how do we know that centroid lies on the line $HC$? $\endgroup$
    – Ekushkebi
    Commented Jun 21, 2015 at 18:59
  • $\begingroup$ @Ekushkebi, done. :) $\endgroup$
    – Apurv
    Commented Jun 21, 2015 at 19:06
  • $\begingroup$ You have the same letter $C$ for two different points. $\endgroup$
    – user26486
    Commented Jun 21, 2015 at 19:46
  • $\begingroup$ @user26486, thanks for pointing out. Rectified! :) $\endgroup$
    – Apurv
    Commented Jun 22, 2015 at 16:15

enter image description here

since $$\angle HAO=|B-C|.OA=R,AH=2R\cos{A}$$ use cosin theorem we have \begin{align*} OH^2&=AH^2+AO^2-2AH\cdot AO\cos{|B-C|} =4R^2\cos^2{A}+R^2-4R^2\cos{A}\cos{(B-C)}\\ &=5R^2-4R^2\sin^2{A}+4R^2\cos{(B+C)}\cos{(B-C)}\\ &=9R^2-4R^2(\sin^2{A}+\sin^2{B}+\sin^2{C})\\ &=9R^2-(a^2+b^2+c^2) \end{align*}

  • $\begingroup$ Can u say why is $ \angle {HAO} \ = \mod{B-C}$ $\endgroup$
    – AbVk1718
    Commented Aug 31, 2021 at 9:19

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