2017 IGO Intermediate P1

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

2017 IGO Intermediate P1

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 4th Iranian Geometry Olympiad (Intermediate) P1

The post below has been deleted. Click to close.

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

178 posts

bgn

#1 h Sep 15, 2017, 6:14 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute-angled triangle with $A=60^{\circ}$ . Let $E,F$ be the feet of altitudes through $B,C$ respectively. Prove that $CE-BF=\tfrac{3}{2}(AC-AB)$

Proposed by Fatemeh Sajadi

Z K Y

The post below has been deleted. Click to close.

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

436 posts

#2 h Sep 15, 2017, 6:51 AM • 2 Y

Y by Adventure10, Mango247

Easy angle/length chasing.

Let $AC=b$ and $AB=c$ .
We have $\angle ABE = 30^{\circ}$ and $\angle ACF = 30^{\circ}$ . Thus, we have $AE = \frac{c}{2}$ and $AF=\frac{b}{2}$ .
Therefore, $CE=b-\frac{c}{2}$ and $BF=c-\frac{b}{2}$ . Subtracting these two yields $CE-BF = \frac{3}{2} b - \frac{3}{2} c = \frac{3}{2} (AC-AB)$ .

Z K Y

The post below has been deleted. Click to close.

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

40 posts

#3 h Oct 8, 2020, 7:29 PM

Y by

Easy Length Chasing Solution :-D

:-D

Solution: Notice that,
$AE = \frac{AB}{2} \text{ [Sine Law]}$ $AF = \frac{AC}{2}$
$AC - AB = (\frac{AB}{2} + CE) - (\frac{AC}{2} + BF)$ $CF - BF = AC - AB - \frac{1}{2}AB + \frac{1}{2}AC = \frac{2AC - 2AB - AB + AC }{2} = \frac{3}{2}(AC - AB)$

This post has been edited 1 time. Last edited by dwip_neel, Oct 8, 2020, 7:30 PM

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

#4 h Dec 27, 2021, 10:52 AM

Y by

CE - BF = (AC-AE) - (AB-AF) = AC - AB/2 - AB + AC/2 = 3/2(AC - AB)

Z K Y

The post below has been deleted. Click to close.

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

lian_the_noob12

173 posts

lian_the_noob12

#5 h Apr 25, 2023, 10:09 AM

Y by

$\color{blue} \boxed{\textbf{SOLUTION}}$

$BE \cap CF \equiv H$

$\color{red} \textbf{Step 1}$
$BF=BHcos30°=\frac{\sqrt 3}{2} BH$

By $\textbf{Similarity}$ ,

$\frac{BF}{BE}=\frac{BH}{AB}$
$\implies AB=\frac{2BE}{\sqrt 3}$

Similarly, $AC=\frac{2CF}{\sqrt 3}$

$\frac{3}{2} (AC-AB) = \sqrt 3 (CF-BE)$

$\color{red} \textbf{Step 2}$
$AE=BEtan30°=\frac{BE}{\sqrt 3}$
$CE=AC-AE= \frac{2CF-BE}{\sqrt 3}$
Similarly,
$BF=\frac{2BE-CF}{\sqrt 3}$

So, $CE-BF= \frac{3CF-3BE}{\sqrt 3}= \sqrt 3(CF-BE)$
Which is desired $\blacksquare$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a