Geometry

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

Geometry

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Belarusian Mathematical Olympiad

The post below has been deleted. Click to close.

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

104 posts

#1 h Mar 31, 2017, 7:25 PM • 1 Y

Y by Adventure10

Let $AA_1, BB_1, CC_1$ be altitudes of an acute-angeled triangle $ABC$ ( $A_1 \in BC, B_1 \in AC, C_1 \in AB$ ). Let $J_a, J_b, J_c$ be centers of inscribed circles of $AC_1B_1$ , $BA_1C_1$ and $CB_1A_1$ respectively. Prove that radius of circumecircle of triangle $J_aJ_bJ_c$ equals radius of inscribed circle of triangle $ABC$

Z K Y

The post below has been deleted. Click to close.

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

776 posts

#2 h Mar 31, 2017, 10:50 PM • 2 Y

Y by Adventure10, Mango247

any solution

Z K Y

The post below has been deleted. Click to close.

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

110 posts

#4 h Apr 1, 2017, 9:35 AM • 1 Y

Y by Adventure10

A formula that we'll use a lot is this: $XI = Y\frac{\sin \frac {1}{2} Z}{\cos \frac {1}{2} Y} = Z\frac{\sin \frac {1}{2} Y}{\cos \frac{1}{2} Z}$ , where I is the incenter of $\Delta XYZ$ . By that, and by noticing that $\Delta A_{1}B_{1}C_{1}$ is the orthic triangle of $\Delta ABC$ , we find that $J_{y}X_{1} = X_{1}Y \frac {\sin \frac{1}{2} Y}{\cos \frac{1}{2} Z} = XI \cos Y$ , using XYZ as any permutation of ABC. Using the law of sines on $\Delta J_{y}X_{1}J_{z}$ , we have that $\frac {\cos Y}{\sin \measuredangle X_{1}J_{z}J_{y}} = \frac {\cos Z}{\sin \measuredangle X_{1}J_{y}J_{z}}$ . We know that $\measuredangle X_{1}J_{z}J_{y} + \measuredangle X_{1}J_{y}J_{z} = X$ So we can write $\frac{\sin (\frac {\pi}{2} - Y)}{\sin (\frac {\pi}{2} - Z)} = \frac{\sin \measuredangle X_{1}J_{z}J_{y}}{\sin \measuredangle X_{1}J_{y}J_{z}}$ and the sum of the angles on both sides is X. Using a lemma, which I will only state here, $\measuredangle X_{1}J_{z}J_{y} = \frac{\pi}{2} - Y$ and $\measuredangle X_{1}J_{y}J_{z} = \frac{\pi}{2} - Z$ . We have, doing this to every angle and noticing that $\measuredangle Z_{1}J_{x}Y_{1} = \frac{\pi + A}{2}$ , that $\measuredangle J_{y}J_{x}J_{z} = \frac{\pi - A}{2}$ . Law of sines again on $\Delta J_{y}X_{1}J_{z}$ , we find that $J_{y}J_{z} = XI \sin X$ . Law of sines on $\Delta J_{x}J_{y}J_{z}$ , $R_{J} = XI \sin \frac {1}{2} X = 4R \sin \frac {1}{2} X \sin \frac {1}{2} Y \sin \frac{1}{2} Z = 4R \sin \frac {1}{2} A \sin \frac {1}{2} B \sin \frac{1}{2} C = r$ , where r is the inradius and R is the circumradius.
Q.E.D

This post has been edited 4 times. Last edited by Domingues3, Apr 1, 2017, 9:40 AM

Z K Y

The post below has been deleted. Click to close.

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

Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#5 h Apr 23, 2022, 7:05 PM

Y by

Note that $AC_1B_1$ and $A_1BC_1$ are similar so $J_bC_1J_a$ and $BC_1B_1$ are similar also $J_aB_1J_c$ and $CB_1C_1$ are similar so
$\angle J_bJ_aJ_c = \frac{\angle B+C}{2}$ . By sines law in $A_1J_bJ_C$ we have $\frac{\sin{A}}{J_bJ_c} = \frac{\sin{90-B}}{A_1J_b}$ Note that $A_1J_b = AI.\cos{B}$ so $J_bJ_c = AI.\sin{A}$ . Let $R$ be circumradius of $J_aJ_bJ_c$ .
Note that $2R = \frac{AI.\sin{A}}{\sin{\frac{B+C}{2}}}$ and $2r = (b+c-a).\sin{\frac{A}{2}}$ so we need to prove $\frac{AI.\sin{A}}{\sin{\frac{B+C}{2}}} = (b+c-a).\sin{\frac{A}{2}}$ or $\frac{\sin{\frac{A}{2}}}{\sin{A}} = \frac{AI}{b+c-a}$ which is true.
we're Done.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a