Cosines Relation with H, O and R - [IMO LongList 1971]

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Cosines Relation with H, O and R - [IMO LongList 1971]

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#1 h Jan 1, 2011, 9:10 PM • 2 Y

Y by Adventure10 and 1 other user

In a triangle $ABC$ , let $H$ be its orthocenter, $O$ its circumcenter, and $R$ its circumradius. Prove that:

(a) $|OH| = R \sqrt{1-8 \cos \alpha \cdot \cos \beta \cdot \cos \gamma}$ where $\alpha, \beta, \gamma$ are angles of the triangle $ABC;$

(b) $O \equiv H$ if and only if $ABC$ is equilateral.

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#2 h Jan 1, 2011, 9:23 PM • 1 Y

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Are there any majot flaws

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#3 h Jan 2, 2011, 8:29 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Proposed problem. Let $\triangle ABC$ with orthocenter $H$ and circumcircle $C(O,R)$ . Show $|OH| = R \sqrt{1-8 \cos A \cdot \cos B \cdot \cos C}$ .

Proof. $OH=3\cdot OG$ and $R^2-OG^2=\frac 19\cdot \left(a^2+b^2+c^2\right)$ $\implies$ $OH^2=9R^2-\sum a^2=$ $9R^2-4R^2\cdot\sum \sin^2A=$ $9R^2-2R^2\cdot\sum (1-\cos 2A)=$ $R^2\left(3+2\cdot\sum\cos 2A\right)$ $\implies \boxed{\ OH^2=R^2\left(3+2\cdot\sum\cos 2A\right)\ }\ (1)$ . Observe that $\sum \cos 2A=\cos 2A+2\cos (B+C)\cos (B-C)=$ $2\cos^2A-1-2\cos A\cos (B-C)=$ $-1-2\cos A[(\cos (B+C)+\os (B-C)]=$ $-1-4\cos A\cos B\cos C$ , i.e. $\boxed{\ \sum\cos 2A=-1-4\cdot \prod\cos A\ }$ . In conclusion, the relation $(1)$ becomes $OH^2=R^2\cdot\left(1-8\cdot\cos A\cos B\cos C\right)$ , i.e. $OH=R\cdot\sqrt {1-8\prod\cos A}$ .
On other hand, $1-8\cdot\prod\cos A=1-4\cos A[\cos (B+C)+\cos (B-C)]=$ $1-4\cos A\cos (B-C)+4\cos^2A=$ $\left[2\cos A-\cos (B-C)\right]^2+\sin^2(B-C)$ . Thus, $1-8\cdot\prod \cos A\ge 0$ , i.e. $\cos A\cos B\cos C\le\frac 18$ . We have the equality $\iff A=B=C\iff O\equiv H$ .

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