300,000 points problem!

Geometry Level 3

In a triangle $ABC$ , we have $\cot A+\cot{B}=\cot{C}$ , where $A$ , $B$ and $C$ are the angles of the triangle and $a$ , $b$ and $c$ are the lengths of the side opposite the angle respectively. Determine $\dfrac{a^2+b^2}{c^2}$ .

The answer is 3.

1 solution

Noel Lo
Jul 6, 2017

$cot{A}+cot{B}=cot{C}$

$\frac{cos{A}}{sin{A}}+\frac{cos{B}}{sin{B}}=\frac{cos{C}}{sin{C}}$

Multiplying throughout by $sin{A}sin{B}sin{C}$ , we have:

$(cos{A}sin{B}+cos{B}sin{A})sin{C}=sin{A}sin{B}cos{C}$

$sin(A+B)sin{C}=sin{A}sin{B}cos{C}$

Considering that $A+B+C=180$ degrees, $sin(A+B)=sin{C}$ therefore:

$sin{C}sin{C}=sin{A}sin{B}cos{C}$

$\frac{sin{C}}{sin{A}}\frac{sin{C}}{sin{B}}=cos{C}$

Employing the sine and cosine rule, we have:

$\frac{c}{a}\frac{c}{b}=\frac{a^2+b^2-c^2}{2ab}$

$\frac{c^2}{ab}=\frac{a^2+b^2-c^2}{2ab}$

Multiplying throughout by $2ab$ ,

$2c^2=a^2+b^2-c^2$

Dividing throughout by $c^2$ :

$2=\frac{a^2+b^2}{c^2}-1$

$\frac{a^2+b^2}{c^2}=2+1$

$\frac{a^2+b^2}{c^2}=\boxed{3}$

@noel lo using the well known identity

cot A + cot B + cot C = ( a^2+b^2+c^2) / (4 delta )

your work reduces significantly and you get a much easier solution

A Former Brilliant Member - 3 years, 11 months ago

300,000 points problem!

The answer is 3.

1 solution

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