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Let O and H be the circumcenter and orthocenter of triangle ABC, respectively. Let a, b, and c denote the side lengths as in the picture below:

Find OH if the circumradius is equal to 7 and a^2 + b^2 + c^2 = 432.

Also, Find AH^2 + BH^2 + CH^2 if the circumradius is equal to 7 and a^2 + b^2 + c^2 = 432.

Aug 4, 2019

#1
+23878
+2

Let O and H be the circumcenter and orthocenter of triangle ABC, respectively.

Let a, b, and c denote the side lengths as in the picture below:

1.
Find OH if the circumradius is equal to 7 and a^2 + b^2 + c^2 = 432.

$$\text{Let circumradius R=7} \\ \text{Let OA=OB=OC=R}$$

$$\vec{OH}=\vec{OA}+\vec{OB}+\vec{OC} \quad \text{This is the Sylvester’s relation}$$

$$\begin{array}{|rcll|} \hline \left(\vec{OH}\right)^2 &=& \left(\vec{OA}+\vec{OB}+\vec{OC}\right)^2 \\ OH^2 &=& \underbrace{OA^2+OB^2+OC^2}_{=3R^2} +2\vec{OA}\vec{OB}+2\vec{OA}\vec{OC}+2\vec{OB}\vec{OC} \\ OH^2 &=& 3R^2 +2\vec{OA}\vec{OB}+2\vec{OA}\vec{OC}+2\vec{OB}\vec{OC} \\\\ && 2\vec{OA}\vec{OB} = 2R^2\ \cos(\angle AOB) & c^2 = 2R^2-2R^2\cos(\angle AOB) \\ &&& 2R^2\cos(\angle AOB)=2R^2-c^2 \\ && \mathbf{2\vec{OA}\vec{OB} = 2R^2-c^2} \\\\ && 2\vec{OA}\vec{OC} = 2R^2\ \cos(\angle AOC) & b^2 = 2R^2-2R^2\cos(\angle AOC) \\ &&& 2R^2\cos(\angle AOC)=2R^2-b^2 \\ && \mathbf{2\vec{OA}\vec{OC} = 2R^2-b^2} \\\\ && 2\vec{OB}\vec{OC} = 2R^2\ \cos(\angle BOC) & a^2 = 2R^2-2R^2\cos(\angle BOC) \\ &&& 2R^2\cos(\angle BOC)=2R^2-a^2 \\ && \mathbf{2\vec{OB}\vec{OC} = 2R^2-a^2} \\\\ OH^2 &=& 3R^2 +2R^2-c^2+2R^2-b^2+2R^2-a^2 \\\\ \mathbf{OH^2} &=& \mathbf{9R^2 -\left(a^2+b^2+c^2\right)} \\ \hline OH^2 &=& 9R^2 -\left(a^2+b^2+c^2\right) \quad | \quad R=7,\ a^2 + b^2 + c^2 = 432 \\ OH^2 &=& 9*7^2 -432 \\ OH^2 &=& 9 \\ \mathbf{OH} &=& \mathbf{3} \\ \hline \end{array}$$

Aug 5, 2019
#2
+23878
+2

Let O and H be the circumcenter and orthocenter of triangle ABC, respectively.
Let a, b, and c denote the side lengths as in the picture below:

2.
Also, Find $$AH^2 + BH^2 + CH^2$$ if the circumradius is equal to 7 and $$a^2 + b^2 + c^2 = 432$$.

$$\begin{array}{|lrcll|} \hline (1) & \vec{OA} + \vec{AH} &=& \vec{OH} \\ (2) & \vec{OB} + \vec{BH} &=& \vec{OH} \\ (3) & \vec{OC} + \vec{CH} &=& \vec{OH} \\ \hline (1)+(2)+(3): & \underbrace{\vec{OA}+\vec{OB}+\vec{OC}}_{=\vec{OH}} +\vec{AH}+\vec{BH}+\vec{CH} &=& 3\vec{OH} \\ & \vec{OH} +\vec{AH}+\vec{BH}+\vec{CH} &=& 3\vec{OH} \\ & \mathbf{\vec{AH}+\vec{BH}+\vec{CH}} &=& \mathbf{2\vec{OH}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \left(\vec{AH}+\vec{BH}+\vec{CH}\right)^2 &=& 4\left(\vec{OH}\right)^2 \\ AH^2+BH^2+CH^2\\ +2\vec{AH}\vec{BH}\\ +2\vec{AH}\vec{CH}\\ +2\vec{BH}\vec{CH}&=& 4\left(\vec{OH}\right)^2 \\\\ && 2\vec{AH}\vec{BH} = 2AH*BH\ \cos(\angle AHB) & c^2 = AH^2+BH^2-2AH*BH\ \cos(\angle AHB) \\ &&& 2AH*BH\ \cos(\angle AHB)=AH^2+BH^2-c^2 \\ && \mathbf{2\vec{AH}\vec{BH} = AH^2+BH^2-c^2} \\\\ \\ && 2\vec{AH}\vec{CH} = 2AH*CH\ \cos(\angle AHC) & b^2 = AH^2+CH^2-2AH*CH\ \cos(\angle AHC) \\ &&& 2AH*CH\ \cos(\angle AHC)=AH^2+CH^2-b^2 \\ && \mathbf{2\vec{AH}\vec{CH} = AH^2+CH^2-b^2} \\\\ \\ && 2\vec{BH}\vec{CH} = 2BH*CH\ \cos(\angle BOC) & a^2 = BH^2+CH^2-2BH*CH\cos(\angle BOC) \\ &&& 2BH*CH\cos(\angle BOC)=BH^2+CH^2-a^2 \\ && \mathbf{2\vec{BH}\vec{CH} = BH^2+CH^2-a^2} \\\\ AH^2+BH^2+CH^2\\ +AH^2+BH^2-c^2\\ +AH^2+CH^2-b^2\\ +BH^2+CH^2-a^2 &=& 4\left(\vec{OH}\right)^2 \\ 3\left( AH^2+BH^2+CH^2\right)-\left(a^2+b^2+c^2\right) &=& 4\left(\vec{OH}\right)^2 \\ 3\left( AH^2+BH^2+CH^2\right) &=& 4\left(\vec{OH}\right)^2+\left(a^2+b^2+c^2\right) \\ 3\left( AH^2+BH^2+CH^2\right) &=& 4*9+432 \\ 3\left( AH^2+BH^2+CH^2\right) &=& 468 \quad | \quad :3 \\ \mathbf{AH^2+BH^2+CH^2} &=& \mathbf{156} \\ \hline \end{array}$$

Aug 5, 2019
#3
+107006
+1

That is very impressive Heueka

Aug 6, 2019
#4
+23878
+2

Thank you, Melody !

heureka  Aug 6, 2019