Triangles

Geometry Level 2

Given that the side lengths $a$ , $b$ , and $c$ opposite $\angle A$ , $\angle B$ , and $\angle C$ respectively of a triangle satisfy the system of equations:

$\begin{cases} c^2 = 2ab \\ a^2 + c^2 = 3b^2 \end{cases}$

What is the measure of $\angle A$ ?

15^\circ

75^\circ

45^\circ

90^\circ

30^\circ

60^\circ

2 solutions

Chris Lewis
Jul 20, 2020

Combining the two given relationships, we get $a^2+2ab=3b^2$ . This is a quadratic; we can rearrange and factorise to get $(a-b)(a+3b)=0$

Now, clearly $a+3b>0$ ; so we find $a=b$ . Substituting back, we get $c=a\sqrt2$ ; so the triangle is right-angled and isosceles, and $A=B=\boxed{45^{\circ}}$ .

nice but it would be easier if you add $b^2$ on both sides

Razing Thunder - 10 months, 3 weeks ago

Good point. I would say factorising (if you spot it) is easier than completing the square; but both are equivalent (we're just solving the quadratic).

Chris Lewis - 10 months, 3 weeks ago

I know right? I did the same.

Manikandan AS - 10 months, 3 weeks ago

Chew-Seong Cheong
Jul 20, 2020

Given that $\begin{cases} \blue{c^2 = 2ab} & ...(1) \\ a^2 + c^2 = 3b^2 & ...(2) \end{cases}$ .

From $(2)$ :

$\begin{aligned} a^2 + \blue{2ab} & = 3b^3 \\ a^2 + 2ab - 3b^3 & = 0 \\ (a+3b)(a-b) & = 0 & \small \blue{\text{Since }a, b > 0} \\ \implies a & = b \end{aligned}$

This means that $\triangle ABC$ is isosceles. From $(2)$ again, $a^2 + c^2 = 3b^2$ , $\implies c^2 = 3b^2 - a^2 = 2b^2 = a^2 + b^2$ , since $a=b$ . Therefore $\triangle ABC$ is also a right triangle at $\angle C$ and $\angle A = \angle B = \boxed{45^\circ}$ .

@Razing Thunder , no need to shout with exclamation mark (!) in the title again. In fact, no punctuation is need in the title.

Chew-Seong Cheong - 10 months, 3 weeks ago

Triangles

2 solutions

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