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.
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+2→OA→OB+2→OA→OC+2→OB→OCOH2=3R2+2→OA→OB+2→OA→OC+2→OB→OC2→OA→OB=2R2 cos(∠AOB)c2=2R2−2R2cos(∠AOB)2R2cos(∠AOB)=2R2−c22→OA→OB=2R2−c22→OA→OC=2R2 cos(∠AOC)b2=2R2−2R2cos(∠AOC)2R2cos(∠AOC)=2R2−b22→OA→OC=2R2−b22→OB→OC=2R2 cos(∠BOC)a2=2R2−2R2cos(∠BOC)2R2cos(∠BOC)=2R2−a22→OB→OC=2R2−a2OH2=3R2+2R2−c2+2R2−b2+2R2−a2OH2=9R2−(a2+b2+c2)OH2=9R2−(a2+b2+c2)|R=7, a2+b2+c2=432OH2=9∗72−432OH2=9OH=3
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=3→OH→OH+→AH+→BH+→CH=3→OH→AH+→BH+→CH=2→OH
(→AH+→BH+→CH)2=4(→OH)2AH2+BH2+CH2+2→AH→BH+2→AH→CH+2→BH→CH=4(→OH)22→AH→BH=2AH∗BH cos(∠AHB)c2=AH2+BH2−2AH∗BH cos(∠AHB)2AH∗BH cos(∠AHB)=AH2+BH2−c22→AH→BH=AH2+BH2−c22→AH→CH=2AH∗CH cos(∠AHC)b2=AH2+CH2−2AH∗CH cos(∠AHC)2AH∗CH cos(∠AHC)=AH2+CH2−b22→AH→CH=AH2+CH2−b22→BH→CH=2BH∗CH cos(∠BOC)a2=BH2+CH2−2BH∗CHcos(∠BOC)2BH∗CHcos(∠BOC)=BH2+CH2−a22→BH→CH=BH2+CH2−a2AH2+BH2+CH2+AH2+BH2−c2+AH2+CH2−b2+BH2+CH2−a2=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)=4∗9+4323(AH2+BH2+CH2)=468|:3AH2+BH2+CH2=156