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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
avatar+26397 
+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.

 

Let circumradius R=7Let OA=OB=OC=R

 

OH=OA+OB+OCThis is the Sylvester’s relation

(OH)2=(OA+OB+OC)2OH2=OA2+OB2+OC2=3R2+2OAOB+2OAOC+2OBOCOH2=3R2+2OAOB+2OAOC+2OBOC2OAOB=2R2 cos(AOB)c2=2R22R2cos(AOB)2R2cos(AOB)=2R2c22OAOB=2R2c22OAOC=2R2 cos(AOC)b2=2R22R2cos(AOC)2R2cos(AOC)=2R2b22OAOC=2R2b22OBOC=2R2 cos(BOC)a2=2R22R2cos(BOC)2R2cos(BOC)=2R2a22OBOC=2R2a2OH2=3R2+2R2c2+2R2b2+2R2a2OH2=9R2(a2+b2+c2)OH2=9R2(a2+b2+c2)|R=7, a2+b2+c2=432OH2=972432OH2=9OH=3

 

 

laugh

 Aug 5, 2019
 #2
avatar+26397 
+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 AH2+BH2+CH2 if the circumradius is equal to 7 and a2+b2+c2=432.

(1)OA+AH=OH(2)OB+BH=OH(3)OC+CH=OH(1)+(2)+(3):OA+OB+OC=OH+AH+BH+CH=3OHOH+AH+BH+CH=3OHAH+BH+CH=2OH

 

(AH+BH+CH)2=4(OH)2AH2+BH2+CH2+2AHBH+2AHCH+2BHCH=4(OH)22AHBH=2AHBH cos(AHB)c2=AH2+BH22AHBH cos(AHB)2AHBH cos(AHB)=AH2+BH2c22AHBH=AH2+BH2c22AHCH=2AHCH cos(AHC)b2=AH2+CH22AHCH cos(AHC)2AHCH cos(AHC)=AH2+CH2b22AHCH=AH2+CH2b22BHCH=2BHCH cos(BOC)a2=BH2+CH22BHCHcos(BOC)2BHCHcos(BOC)=BH2+CH2a22BHCH=BH2+CH2a2AH2+BH2+CH2+AH2+BH2c2+AH2+CH2b2+BH2+CH2a2=4(OH)23(AH2+BH2+CH2)(a2+b2+c2)=4(OH)23(AH2+BH2+CH2)=4(OH)2+(a2+b2+c2)3(AH2+BH2+CH2)=49+4323(AH2+BH2+CH2)=468|:3AH2+BH2+CH2=156

 

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 Aug 5, 2019
 #3
avatar+118694 
+1

That is very impressive Heueka      cool

 Aug 6, 2019
 #4
avatar+26397 
+2

Thank you, Melody !

 

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heureka  Aug 6, 2019

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