distances of a center in circle from sides of isosceles triangle

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

distances of a center in circle from sides of isosceles triangle

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 2006 Spanish Mathematical Olympiad P3

The post below has been deleted. Click to close.

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

30629 posts

#1 h Jul 20, 2018, 9:26 PM • 1 Y

Y by Adventure10

$ABC$ is an isosceles triangle with $AB = AC$ . Let $P$ be any point of a circle tangent to the sides $AB$ in $B$ and to AC in C. Denote $a$ , $b$ and $c$ to the distances from $P$ to the sides $BC, AC$ and $AB$ respectively. Prove that: $a^2=bc$

This post has been edited 1 time. Last edited by parmenides51, Jul 20, 2018, 9:52 PM
Reason: corrected problem number

Z K Y

The post below has been deleted. Click to close.

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

9029 posts

#2 h Aug 24, 2018, 6:53 PM • 2 Y

Y by Adventure10, Mango247

Solution

Z K Y

The post below has been deleted. Click to close.

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

1585 posts

#3 h Aug 25, 2018, 2:37 PM • 3 Y

Y by parmenides51, Adventure10, Mango247

The problem is much older than 2006. There is in a greek book (Γεωμετρία γιάννη ντάνη) before 1972. It is the proposition 714 in page 267.

Attachments:

This post has been edited 3 times. Last edited by george_54, Aug 25, 2018, 3:36 PM

Z K Y

The post below has been deleted. Click to close.

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

527 posts

#4 h Aug 25, 2018, 11:50 PM • 2 Y

Y by Adventure10, Mango247

Dear George,

how does Tanis prove this?

Friendly,
M.T.

Z K Y

The post below has been deleted. Click to close.

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

30629 posts

#5 h Mar 9, 2019, 10:29 PM • 2 Y

Y by Adventure10, Mango247

This is Ntanis' proof (the green greek book), he uses similarity

Attachments:

This post has been edited 1 time. Last edited by parmenides51, Mar 9, 2019, 10:30 PM

Z K Y

The post below has been deleted. Click to close.

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

1618 posts

#6 h Mar 10, 2019, 12:52 PM • 1 Y

Y by Adventure10

Let $D$ , $E$ , $F$ are orthogonal projections of $P$ on $BC$ , $CA$ , $AB$
We have: $\widehat{FPD}$ = $180^o$ $-$ $\widehat{FBD}$ = $180^o$ $-$ $\widehat{DCE}$ = $\widehat{DPF}$ and $\widehat{PFD}$ = $\widehat{PBD}$ = $\widehat{PCE}$ = $\widehat{PDE}$ then: $\triangle$ $PFD$ $\sim$ $\triangle$ $PDE$
So: $\dfrac{PF}{PD}$ = $\dfrac{PD}{PE}$ or $PD^2$ = $PE$ . $PF$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a