3A + 2B = 180° ==> a^2 + bc = c^2

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3A + 2B = 180° ==> a^2 + bc = c^2

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Source: Bundeswettbewerb Mathematik 2005, 1st Round, problem 3

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25 posts

#1 h Mar 1, 2005, 11:04 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle with sides $a$ , $b$ , $c$ and (corresponding) angles $A$ , $B$ , $C$ .
Prove that if $3A + 2B = 180^{\circ}$ , then $a^2+bc=c^2$ .

Additional problem:
Prove that the converse also holds, i. e. prove the following:
Let $ABC$ be an arbitrary triangle. Then, $3A + 2B = 180^{\circ}$ if and only if $a^2+bc=c^2$ .

Similar problem:
Let $ABC$ be an arbitrary triangle. Then, $3A + 2B = 360^{\circ}$ if and only if $a^2-bc=c^2$ .

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256 posts

Leon

#2 h Mar 2, 2005, 11:46 AM • 2 Y

Y by Adventure10, Mango247

Theorem 1. Let $\triangle ABC$ be an arbitrary triangle. Then $3A + 2B = 180^\circ$ if and only if $a^2 + bc = c^2$

Proof.

Since $A + B + C = 180^\circ$ and $a = 2R\sin \alpha \ \ ,b = 2R\sin \beta ,\ \ c = 2R\sin \gamma$ we have

$3A + 2B = 180^\circ \Leftrightarrow$

$3A + 2B + C = A + B + C \Leftrightarrow$

$2A + B = C \Leftrightarrow$

$\sin \left( {2A + B} \right) = \sin C \Leftrightarrow$

$\sin B \cdot \sin \left( {2A + B} \right) = \sin B \cdot \sin C \Leftrightarrow$

$\left( {2\sin \frac{B}{2}\cos \frac{B}{2}} \right)\left( {2\sin \frac{{2A + B}}{2}\cos \frac{{2A + B}}{2}} \right) = \sin B \cdot \sin C \Leftrightarrow$

$\left( {2\cos \frac{{2A + B}}{2}\sin \frac{B}{2}} \right)\left( {2\sin \frac{{2A + B}}{2}\cos \frac{B}{2}} \right) = \sin B \cdot \sin C \Leftrightarrow$

$\left[ {\sin \left( {A + B} \right) - \sin A} \right]\left[ {\sin \left( {A + B} \right) + \sin A} \right] = \sin B \cdot \sin C \Leftrightarrow$

$\sin ^2 \left( {A + B} \right) - \sin ^2 A = \sin B \cdot \sin C \Leftrightarrow$

$\sin ^2 C = \sin ^2 A + \sin B \cdot \sin C \Leftrightarrow$

$4R^2 \sin ^2 C = 4R^2 \sin ^2 A + 4R^2 \sin B \cdot \sin C \Leftrightarrow$

$c^2 = a^2 + bc$

and we are done. $\blacksquare$

Theorem 2. Let $\triangle ABC$ be an arbitrary triangle. Then $3A + 2B = 360^\circ$ if and only if $a^2 - bc = c^2$

Proof.

Since $A + B + C = 180^\circ$ and $a = 2R\sin \alpha \ \ ,b = 2R\sin \beta ,\ \ c = 2R\sin \gamma$ we have

$3A + 2B = 360^\circ \Leftrightarrow$

$3\left( {180^\circ - B - C} \right) + 2B = 360^\circ \Leftrightarrow$

$B + 3C = 180^\circ \Leftrightarrow$

$B + 2C = 180^\circ - C \Leftrightarrow$

$\sin \left( {B + 3C} \right) = \sin C \Leftrightarrow$

$\sin B \cdot \sin \left( {B + 3C} \right) = \sin B \cdot \sin C \Leftrightarrow$

$\left( {2\sin \frac{B}{2}\cos \frac{B}{2}} \right)\left( {2\sin \frac{{B + 3C}}{2}\cos \frac{{B + 3C}}{2}} \right) = \sin B \cdot \sin C \Leftrightarrow$

$\left( {2\cos \frac{{B + 3C}}{2}\sin \frac{B}{2}} \right)\left( {2\sin \frac{{B + 3C}}{2}\cos \frac{B}{2}} \right) = \sin B \cdot \sin C \Leftrightarrow$

$\left[ {\sin \left( {B + C} \right) - \sin C} \right]\left[ {\sin \left( {B + C} \right) + \sin C} \right] = \sin B \cdot \sin C \Leftrightarrow$

$\sin ^2 \left( {B + C} \right) - \sin ^2 C = \sin B \cdot \sin C \Leftrightarrow$

$\sin ^2 A = \sin ^2 C + \sin B \cdot \sin C \Leftrightarrow$

$4R^2 \sin ^2 A = 4R^2 \sin ^2 C + 4R^2 \sin B \cdot \sin C \Leftrightarrow$

$a^2 = c^2 + bc \Leftrightarrow$

$a^2 - bc = c^2 \ \ \ \blacksquare$

Remark. In the proof of theorem 1 we have

$2A + B = C \Leftrightarrow \sin \left( {2A + B} \right) = \sin C$

since, if it were $2A + B +C = 180^\circ$ , it would have $A = 0$ .

Similarly, in the proof of theorem 2 we have

$B + 2C = 180^\circ - C \Leftrightarrow \sin \left( {B + 3C} \right) = \sin C$

since, if it were $B +3C = C$ it would have $B = 0, C = 0$ .

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enescu

#3 h Mar 4, 2005, 11:10 AM • 3 Y

Y by Adventure10, sabkx, Mango247

The condition $3A+2B=180^\circ$ is equivalent to $C=90^\circ +\frac A2$ . Therefore $C$ is obtuse. Take $D$ on $AB$ such that $AD=AC$ . Then $\angle BCD=C-\angle ACD=C-\frac {180^\circ -A}{2}=A$ . It follows that triangles $ACB$ and $CDB$ are similar, hence $\frac ac=\frac {c-b}{a}$ , so $a^2+bc=c^2$ .

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Myth

#4 h Mar 4, 2005, 11:52 AM • 2 Y

Y by Adventure10, Mango247

BTW, this problem was discussed it least twice on forum.

Unfortunately, I can't find the links. Ask Darij

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#5 h Mar 4, 2005, 12:57 PM • 2 Y

Y by Adventure10, Mango247

I remember only the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=4230 , where it was shown that if $a^2=c\left(c+b\right)$ (what is equivalent to $a^2-bc=c^2$ ), then A = 2C (what is equivalent to 3A + 2B = 360°). The converse was shown in http://www.mathlinks.ro/Forum/viewtopic.php?t=45481 post #7.

But indeed, the synthetic solution of the initial problem given by Enescu in post #3 can be also used for proving the converse, and the related problem can be solved absolutely similarly.

darij

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#6 h Jul 22, 2023, 10:22 AM

Y by

So 1 St part,if you solve a little bit,we get
C=90°+ (A/2)
sin C=sin{90°+ (A/2)}
sin C=cos(A/2)
Since, Sin C=(c/a)*sinA=2(c/a)*sin(A/2)*cos(A/2)
SO, sin (A/2)=a/2c
-cos C=a/2c
(c²-a²-b²)/2ab=a/2c
So, if we solve we get,
(c+b){c²-bc-a²}=0
Since c≠b
c²-bc-a²=0,a²+bc=c²( proved)
The converse can also be done some way ,just back calculation
Additional problem:
So if we solve we shall get,
(A/2)=C
Then,
cos(A/2)=a/2c
cos C=a/2c
So if we solve,
( c-b){a²-c²-bc}=0
c-b=0. or. a²-bc=c²( proved)
c=b
# KRISHIJIVI

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