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Source: INMO 1992 Problem 1

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1592 posts

Rushil

#1 h Oct 3, 2005, 11:58 PM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC$ , $\angle A = 2 \cdot \angle B$ . Prove that $a^2 = b (b+c)$ .

This post has been edited 1 time. Last edited by Rushil, Oct 4, 2005, 5:39 AM

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3498 posts

#2 h Oct 4, 2005, 2:05 AM • 2 Y

Y by Adventure10, Mango247

Rushil wrote:

INMO 1992 Problem 1

In a triangle $ABC$ , $\angle A = 2 \times \angle B$ . Prove that $a^2 = b (b+c)$

Try to search for Double Angle Theorem.

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578 posts

#3 h Dec 20, 2005, 6:20 PM • 3 Y

Y by AmitayasB, Adventure10, Mango247

$\frac{a}{\sin 2x} = \frac{b}{\sin x} = \frac{c}{\sin 3x}$ $\Longrightarrow$ $c = b(4\cos^2 x - 1)$ and $a = 2b\cos x$ .
$b(b + c) = 4b^2\cos^2 x = a^2$

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146 posts

#4 h Dec 20, 2005, 8:33 PM • 1 Y

Y by Adventure10

Andreas wrote:

$\frac{a}{\sin 2x} = \frac{b}{\sin x} = \frac{c}{\sin 3x}$

How did you get this one?

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1444 posts

#5 h Dec 20, 2005, 8:43 PM • 1 Y

Y by Adventure10

That's just the law of sines

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1421 posts

#6 h Dec 20, 2005, 8:43 PM • 1 Y

Y by Adventure10

$\sin{C}=\sin{(180-A-B)}=\sin{(180-3x)}=\sin{(3x)}$ , Because $\sin{x}=\sin{(180-x)}$ .

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The QuattoMaster 6000

1184 posts

The QuattoMaster 6000

#7 h Jan 17, 2008, 12:02 AM • 1 Y

Y by Adventure10

Rushil wrote:

In a triangle $ABC$ , $\angle A = 2 \cdot \angle B$ . Prove that $a^2 = b (b + c)$ .

Sorry for awakening an old topic, but here's a neat solution that no-one here mentioned already:
Solution

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99 posts

earth

#8 h Apr 19, 2010, 7:04 PM • 3 Y

Y by Samujjal101, Adventure10, Mango247

Hi,here is a good one ! please try out!

Click to reveal hidden text

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Mateescu Constantin

1842 posts

Mateescu Constantin

#9 h Apr 20, 2010, 8:59 PM • 2 Y

Y by Adventure10, Mango247

Another solution

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154 posts

#10 h Jan 25, 2012, 3:22 AM • 5 Y

Y by SHREYAS333, Wizard_32, PME2018, Adventure10, and 1 other user

A pure geometric solution

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2 posts

#11 h Sep 1, 2018, 5:04 AM • 3 Y

Y by sayanjoddar, Adventure10, Mango247

Sorry for reviving a very old thread !
We have to prove a^2=b(b+c)
Rearranging the terms we get a/b=(b+c)/a
L.H.S:-
By sine rule, a/b=sinA/sinB=2cosB ...(A=2B)
R.H.S:-
By sine rule, (b+c)/a=(sinB+sinC)/sinA=(2sin2B*cosB)/sin2B=2cosB ...(A=2B)
Hence, L.H.S=R.H.S proved.

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PhysicsMonster_01

1445 posts

PhysicsMonster_01

#12 h Sep 13, 2018, 7:47 AM • 4 Y

Y by pikazag, MathematicalGiant, Adventure10, Mango247

A bit different solution with the use of trigonometry

This post has been edited 2 times. Last edited by PhysicsMonster_01, Sep 13, 2018, 7:48 AM
Reason: LaTeX

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LoveMaths26102003

84 posts

LoveMaths26102003

#13 h Sep 13, 2018, 8:03 AM • 2 Y

Y by Adventure10, Mango247

Similarity

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651 posts

#14 h Feb 15, 2020, 12:37 PM • 2 Y

Y by Adventure10, Mango247

Suppose, $\angle B =b'$ , $\implies \angle A =2b'$ ,
and, $\angle C =180-3b'$ .

See that, $\frac{C}{\sin 3b'} =\frac{b}{\sin b'}$ .

$\implies c=b(3-4\sin ^2 b')$ .

$\implies c=b(2\cos2b'+1)$ .

$\implies c^2 = bc+2bc \cos 2b'$ .

Using cosine rule we have ,

$a^2=b^2+c^2 -2bc \cos 2b'$ .

$\implies a^2=b^2 +bc$ .

As desired. $\boxed{a^2 = b (b+c)}$

This post has been edited 1 time. Last edited by ftheftics, Feb 15, 2020, 12:38 PM
Reason: BNNn

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41471 posts

sqing

#15 h Feb 17, 2020, 2:19 AM • 1 Y

Y by Adventure10

Rushil wrote:

In a triangle $ABC$ , $\angle A = 2 \angle B$ . Prove that $a^2 = b (b+c)$ .

$\frac{a}{\sin 2B} = \frac{b}{\sin B}\Leftrightarrow \frac{a}{ 2b}=cosB=\frac{c^2+a^2-b^2}{ 2ca} \Leftrightarrow a^2 = b (b+c)$
In a triangle $ABC$ , $a^2 = b (b+c)$ . Prove that $\angle A = 2\angle B.$

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1119 posts

#16 h Feb 17, 2020, 3:03 AM • 1 Y

Y by Adventure10

mathmanman wrote:

That's just the law of sines

oops that link failed

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41471 posts

sqing

#17 h Feb 17, 2020, 12:20 PM • 1 Y

Y by Adventure10

BestChoice123 wrote:

mathmanman wrote:

That's just the law of sines

oops that link failed

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209 posts

S1281

#18 h Apr 19, 2021, 12:12 PM

Y by

We know that,
$a/sinA=b/sinB=c/sinC$
$a/sin2x=b /sinx=c/sin(180^{\circ{}}-3x)$
$a/2cosx=b$
$c/sin3x=b/sinx$
$c/(3sinx-4sin^3x)=b/sinx$
$c/(3-4sin^2x)=b$
$b(b+c)=b^2+bc=\frac{a^2}{4cos^2x}+\frac{a}{2cosx}(3-4sin^2x)\frac{a}{2cosx}=\frac{a^2}{4 cos^2x}4(1-sin^2x)=\frac{a^2}{4 cos^2x}4cos^2x=a^2$ .

This post has been edited 2 times. Last edited by S1281, Apr 19, 2021, 12:13 PM

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456 posts

#19 h May 14, 2022, 6:55 PM

Y by

I was just about the bash thing with trigonometry. But here is a geometric solution I guess.

Solution: Denote $D$ to be the foot of angle bisector of $\angle BAC$ on side $BC$ . Now it is easy to notice that $\angle CAD = \angle ABD$ . So we can deduce that $AC$ is tangent to $\odot(ADB)$ . Applying Power of a Point along with Angle Bisector Theorem we get
$\begin{align*} b^2= \frac{ab}{b+c} \cdot a \\ \implies a^2= b(b+c) \end{align*}$ which proves the desired. $\blacksquare$

This post has been edited 2 times. Last edited by SatisfiedMagma, May 14, 2022, 6:57 PM

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#20 h Aug 6, 2023, 4:59 AM

Y by

Let angle B=x
Then , angle A=2x
By sine rule,
a/ b=sin 2x/ sin x
cos x =a/2b
cos x=c²+a²-b²/2ac
bc²+ba²-b²-a²c=0
(c-b){b(b+c)-a²}=0
If c=b
Then also a²=b(b+c)
If b(b+c)-a²=0
a²=b(b+c)
@ Krishijivi

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