We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
149
4
avatar

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
avatar+23293 
+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.

 

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

 

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

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

 

 

laugh

 Aug 5, 2019
 #2
avatar+23293 
+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 \(a^2 + b^2 + c^2 = 432\).

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

 

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

 

laugh

 Aug 5, 2019
 #3
avatar+105540 
+1

That is very impressive Heueka      cool

 Aug 6, 2019
 #4
avatar+23293 
+2

Thank you, Melody !

 

laugh

heureka  Aug 6, 2019

10 Online Users

avatar
avatar