# Distance between orthocenter and circumcenter.

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.

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

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.

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

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*}

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