Some Identity that I need help

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Some Identity that I need help

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

151 posts

#1 h Dec 28, 2024, 3:41 AM

Y by

Given $\triangle ABC$ with orthocenter , circumcenter and incenter $H,O,I$ , circum-radius $R$ , in-radius $r$ .
Prove that $OH^2 = 2 HI^2 - 4r^2 + R^2$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

151 posts

#2 h Jan 6, 2025, 9:17 AM

Y by

Bump Bump

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

33 posts

Tkn

#3 h Apr 10, 2025, 7:50 AM

Y by

To show the above equality, we need to prove these two claims:

Claim 1. $OH^2=9R^2-(a^2+b^2+c^2)$
Proof. This one is easy by vector bash with the Euler line. Note that
$\frac{OG}{OH}=\frac{1}{3}$ where $G$ denotes the centroid of $\triangle{ABC}$ . It is easy to see that
$\overrightarrow{OH}=3\overrightarrow{OG}=3\left(\frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}\right).$ Comparing modulus of both sides:
$\begin{align*} |\overrightarrow{OH}|^2&=\overrightarrow{OH}\cdot \overrightarrow{OH}\\ &=3R^2+2R^2\left(\cos(2\angle{A})+\cos(2\angle{B})+\cos(2\angle{C})\right)\\ &=3R^2+6R^2-4R^2\left(\sin^2(\angle{A})+\sin^2(\angle{B})+\sin^2(\angle{C})\right)\\ &=9R^2-(a^2+b^2+c^2) \end{align*}$ The last step requires the sine's law on $\triangle{ABC}$ , as in the form:
$\frac{a}{\sin(\angle{A})}=\frac{b}{\sin(\angle{B})}=\frac{c}{\sin(\angle{C})}=2R.$ where $a,b$ and $c$ are side lengths of $BC,CA$ and $AB$ respectively.

Claim 2. $HI^2=2r^2-4R^2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})$
Proof. Given that $M$ is a midpoint of the segment $\overline{BC}$ .
We use the well-known lemma to find, $AH=2OM=2R\cos(\angle{A})$ . And,
$AI=2r\cos\left(\frac{\angle A}{2}\right)=4R\sin\left(\frac{\angle{B}}{2}\right)\sin\left(\frac{\angle{C}}{2}\right).$ The rest is just some bashing of cosine law:
$\begin{align*} HI^2&=AI^2+AH^2-2AI\cdot AH\cos\left(\frac{\angle{B}-\angle{C}}{2}\right)\\ &=16R^2\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)+4R^2\cos^2(\angle{A})-16R^2\cos(\angle{A})\sin\left(\frac{\angle{B}}{2}\right)\sin\left(\frac{\angle{C}}{2}\right)\cos\left(\frac{\angle{B}-\angle{C}}{2}\right)\\ &=4R^2\left(\cos^2(\angle{A})+(4-4\cos(\angle{A}))\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)-\cos(\angle{A})\sin(\angle{B})\sin(\angle{C})\right)\\ &=4R^2\left(\cos^2(\angle{A})+8\sin^2\left(\frac{\angle{A}}{2}\right)\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)-\cos(\angle{A})\sin(\angle{B})\sin(\angle{C})\right)\\ &=4R^2\left(8\sin^2\left(\frac{\angle{A}}{2}\right)\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)+\cos(\angle{A})\left(\cos(\angle{A})-\sin(\angle{B})\sin(\angle{C})\right)\right)\\ &=4R^2\left(8\sin^2\left(\frac{\angle{A}}{2}\right)\sin^2\left(\frac{\angle{B}}{2}\right)\sin^2\left(\frac{\angle{C}}{2}\right)-\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})\right)\\ &=2r^2-4R^2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C}) \end{align*}$ Which is complete.

Then, it just equivalent of showing:
$8R^2(1+\cos(\angle{A})\cos(\angle{B})\cos(\angle{C}))=a^2+b^2+c^2$ Note that $a^2=4R^2\sin^2(\angle{A})$ , similarly for $b$ and $c$ . Simplify and gives
$2\left(1+\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})\right)=\sin^2(\angle{A})+\sin^2{(\angle{B})}+\sin^2(\angle{C})$ or equivalent to
$1=2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})+\cos^2(\angle{A})+\cos^{2}(\angle{B})+\cos^2(\angle{C})$ This is obviously true due to
$\begin{align*} 2\cos(\angle{A})\cos(\angle{B})\cos(\angle{C})&=-\cos^2(\angle{C})+\cos(\angle{A}-\angle{B})\cos(\angle{C})\\ &=-\cos^2(\angle{C})+\frac{1}{2}\left(\cos(\angle{A}-\angle{B}+\angle{C})+\cos(\angle{A}-\angle{B}-\angle{C})\right)\\ &=-\cos^2(\angle{C})+\frac{1}{2}(\cos(180^{\circ}-2B)+\cos(180^{\circ}-2A))\\ &=-\cos^2(\angle{C})-\frac{1}{2}(\cos(2\angle{A})+\cos(2\angle B))\\ &=-\cos^2(\angle{C})-\cos^2(\angle{A})-\cos^2(\angle{B})+1 \end{align*}$ Which actually solves the problem.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a