find the triangle with R(b+c)=a\sqrt{bc}

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

find the triangle with R(b+c)=a\sqrt{bc}

G H J

Reveal topic tags

trig identities

L

G H BBookmark kLocked kLocked NReply

Source: JBMO 1997, Problem 4

The post below has been deleted. Click to close.

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

Valentin Vornicu

7301 posts

Valentin Vornicu

#1 h Oct 30, 2005, 1:19 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Determine the triangle with sides $a,b,c$ and circumradius $R$ for which $R(b+c) = a\sqrt{bc}$ .

Romania

Z K Y

The post below has been deleted. Click to close.

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

2105 posts

#2 h Oct 30, 2005, 2:13 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Valentin Vornicu wrote:

Determine the triangle with sides $a,b,c$ and circumradius $R$ for which $R(b+c) = a\sqrt{bc}$ .

Romania

$\frac{b+c}{\sqrt{bc}}\geq 2$
while $a\leq 2R$ (chord-diameter)

so $b=c$ and $A=90^0$

Z K Y

The post below has been deleted. Click to close.

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

639 posts

M4RI0

#3 h Feb 18, 2006, 9:49 PM • 1 Y

Y by Adventure10

$a\geq 2R$ (chord-diameter)

so $b=c$ and $A=90^0$

Z K Y

The post below has been deleted. Click to close.

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

1828 posts

José

#4 h Feb 19, 2006, 6:08 PM • 2 Y

Y by Adventure10 and 1 other user

socrates wrote $a\leq2R$ .
M4RIO wrote $a\geq2R$ .

Who's wrong???

Z K Y

The post below has been deleted. Click to close.

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

2105 posts

#5 h Feb 20, 2006, 9:34 PM • 2 Y

Y by Adventure10, Mango247

José wrote:

socrates wrote $a\leq2R$ .
M4RIO wrote $a\geq2R$ .

Who's wrong???

Try this by yourself. Do some sketch....
I think i am not the liar..

Z K Y

The post below has been deleted. Click to close.

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

1677 posts

#6 h Feb 20, 2006, 11:50 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Click to reveal hidden text

Z K Y

The post below has been deleted. Click to close.

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

6194 posts

#7 h Feb 21, 2006, 12:14 AM • 2 Y

Y by Adventure10, Mango247

José wrote:

socrates wrote $a\leq2R$ .
M4RIO wrote $a\geq2R$ .

Who's wrong???

obviously $a \le 2R$ is correct because if you have a circle, the largest chord is the diameter

Z K Y

The post below has been deleted. Click to close.

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

547 posts

#8 h Apr 23, 2011, 2:33 AM • 3 Y

Y by RockmanEX3, Adventure10, and 1 other user

This thing follows directly from AM-GM:
$R(b+c)>=2R\sqrt{bc}>=a\sqrt{bc}$ .
This means that $a=2R$ which means that $a$ is the diameter. Furthermore, the triangle is right. Also, equality holds in AM-GM when all the terms are equal so $b=c$ . Therefore the triangle is isosceles right triangle which is of the for $(a,a,a\sqrt{2})$ .
Nice problem.

Z K Y

The post below has been deleted. Click to close.

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

803 posts

#9 h Oct 20, 2012, 4:40 PM • 1 Y

Y by Adventure10

using sine rule yields ,
$sinB+sinC=2sinA\sqrt{sinBsinC}\le 2\sqrt{sinBsinC}$
on the other hand , $sinB+sinC\ge 2\sqrt{sinBsinC}$
so , $sinB+sinC=2\sqrt{sinBsinC} \implies B=C$ and $A=\pi/2$

Z K Y

The post below has been deleted. Click to close.

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

318 posts

#10 h Oct 21, 2012, 12:57 PM • 1 Y

Y by Adventure10

We know $R^2=a^2bc/(b+c)^2$
We have $R^2=a^2/4sin^2A$ thus $sin^2A=(b+c)^2/4bc$ ≥ $1$
also we have $1$ ≥ $sin^2A$
thus $A=90$ and $b=c$

Z K Y

The post below has been deleted. Click to close.

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

Dr Sonnhard Graubner

16100 posts

Dr Sonnhard Graubner

#11 h Oct 21, 2012, 7:23 PM • 2 Y

Y by Adventure10, Mango247

hello, with
$R=\frac{abc}{4A}$
we get
$\frac{(b+c)bc}{4\sqrt{bc}}=A$
$\frac{b+c}{2\sqrt{bc}}=\sin(\alpha)\le 1$ thus
we obtain
$b+c\le 2\sqrt{bc}$ and we get
$b=c$ and $\alpha=90^{\circ}$
Sonnhard.

Z K Y

The post below has been deleted. Click to close.

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

196 posts

#12 h Jul 26, 2014, 4:09 PM • 2 Y

Y by Adventure10, Mango247

(a*√(b*c))/(b+c)=(a*b*c)/4S=(a*b*c)/(2*b*c*sinA)→sin A=(b+c)/(2*√(2&b*c))>=1 and sin A<=1→sin A=1→m(<A)=90 and b=c

Z K Y

The post below has been deleted. Click to close.

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

117 posts

#13 h May 19, 2015, 4:12 PM • 2 Y

Y by Adventure10, Mango247

Valentin Vornicu wrote:

Determine the triangle with sides $a,b,c$ and circumradius $R$ for which $R(b+c) = a\sqrt{bc}$ .

Romania

$a+b \geq 2 \sqrt{bc}$ we get $a\sqrt{bc} \geq 2R\sqrt{bc}$ => $a \geq 2R$ but $a \leq 2R$ and we get $a = 2R$ and $b = c$ .

This post has been edited 1 time. Last edited by neverlose, May 19, 2015, 4:12 PM

Z K Y

The post below has been deleted. Click to close.

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

3 posts

#14 h Jan 16, 2025, 1:37 PM

Y by

we know that by law of sines R = a / 2sinA, so a * sqr(bc)/(b+c) = a/2sinA, dividing by a we have sinA = (b+c)/[2sqr(bc)] = 1/2(sqr(b/c)+sqr(c/b)) and MA => MG inequality we have 1/2(sqr(b/c)+sqr(c/b)) => 1 but sinA is less or equal to 1 so sinA=1 and A=90º. now we know R = a/2 and now substituting in the equation we have b=c so B = C = 45º
FINAL ANSWER : A = 90º, B = C = 45º ( this problem is also 2013 Brazil Cono Sur Training List 4.3 )

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a