The sides a,b,c of a $\Delta$ ABC are in Geometric Progression whose common ratio is $\frac{2}{3}$ and the circumradius of the triangle is $6 \times \sqrt{ \frac{7}{209}}$ .

Find the longest side of the triangle .

The answer is of the form $\frac{m}{n}$ where m and n are co-prime numbers ;

Find $m^{n}$ .

You can try more of my Questions here .

The answer is 2401.

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Actually this the way I did it .

Happy to see that there is someone who thinks like me !!!!

A Former Brilliant Member
- 6 years, 5 months ago

Hi, its been a long time since I solved a question of yours what happened , can you post one sooner?

I'll be waiting !!!

A Former Brilliant Member
- 6 years, 4 months ago

Let the longest side length be $a$ , $b=\frac{2}{3}a$ , $c=\frac{4}{9}a$ , the center of the circumcircle be $O$ , its radius $r = 6\times \sqrt{\frac{7}{209}}$ , $\angle ABO = \alpha$ and $\angle OBC = \beta$ .

Using Cosine Rule, we have:

$\begin{cases} b^2 = a^2 + c^2 - 2 ac \cos{(\alpha+\beta)} &\Rightarrow \cos{(\alpha+\beta)} = \dfrac {61}{72} &...(1)\\ r^2 = c^2 + r^2 - 2 cr \cos{\alpha} &\Rightarrow \cos{\alpha} = \dfrac {2a}{9r} &...(2)\\ r^2 = a^2 + r^2 - 2 ar \cos{\beta} &\Rightarrow \cos{\beta} = \dfrac {a}{2r} &...(3) \end{cases}$

We know that:

$\cos{(\alpha+\beta)} = \cos{\alpha} \cos{\beta} - \sin{\alpha} \sin{\beta}$

$\displaystyle \Rightarrow \frac {61}{72} = \frac {2a}{9r} \times \dfrac {a}{2r} - \sqrt{\left(1-\frac {4a^2}{81r^2}\right) \left(1-\frac {a^2}{4r^2}\right)}$

$\displaystyle \Rightarrow \frac {61}{72} - \frac {a^2}{9r^2} = - \sqrt{\left(1-\frac {4a^2}{81r^2}\right) \left(1-\frac {a^2}{4r^2}\right)}$

$\displaystyle \Rightarrow \frac {3721}{5184} - \frac {61a^2}{324r^2} + \frac {a^4}{81r^4} = 1-\frac {97a^2}{324r^2} + \frac {a^4}{81r^4}$

$\Rightarrow \dfrac {a^2}{9r^2} = \dfrac {1463}{5184} \quad \Rightarrow a^2 = \dfrac {1463\dot{} 9\dot{} 36 \dot{} 7}{5184\dot{} 209} = \dfrac {49}{16}$

$\Rightarrow a = \dfrac {7}{4}\quad \Rightarrow m^n = 7^4 = \boxed{2401}$

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Nice solution but there was an easy way R=(abc)/(4*Area),where you apply Heron's formula for the area of the triangle

Rifath Rahman
- 6 years, 5 months ago

Nice solution Sir

A Former Brilliant Member
- 6 years, 5 months ago

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Nice question, actually.

Chew-Seong Cheong
- 6 years, 5 months ago

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Thanks sir.

Sir, from where did you practice Geometry since it's almost as if it's your strong point ,right ?

A Former Brilliant Member
- 6 years, 5 months ago

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@A Former Brilliant Member – I am a retired person and I have a lot of time practicing math on Brilliant.org. Just digging through what I have learned 30 years ago and start practicing. Geometry hasn't changed much over the years. In fact, I didn't learn about incircle, circumcircle, centroid back then. I just picked up through the internet. Work hard and you will be as good if not better.

Chew-Seong Cheong
- 6 years, 5 months ago

$R=\dfrac{product\ of\ sides}{4*Area}.\\
Let \ the\ sides\ be\ 18x,\ 12x,\ and\ 8x.\\
\therefore\ s=19x,\ \ \ \ \ Area=\sqrt{19*1*7*11}*x^2.\\
\implies\ \dfrac{18*12*8*x^3}{4* \sqrt{19*1*7*11}*x^2}=\dfrac{12^3x}{4*\sqrt{7*209}}=R=6*\sqrt{\dfrac 7 {209}}.\\
So\ x=\dfrac 7 {72}.\\ \therefore\ Biggest\ side\ =18x=\dfrac 7 4=\dfrac m n.\\
So\ \ m^n=7^4=\color{#D61F06}{2401}$
.

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We have a triangle of sides $a$ , $\frac{2}{3}a$ and $\frac{4}{9}a$ , inscribed within a circle of radius $\sqrt{ \frac{252}{209}}$ .

By the Cosine Rule, $a^2 = (\frac{2}{3}a)^2 + (\frac{4}{9}a)^2 - 2 \times \frac{2}{3}a \times \frac{4}{9}a \times \cos\theta$ , where $\theta$ is the angle opposite $a$ ,

$\implies \frac{29}{81}a^2 = -\frac{16}{27}a^2\cos\theta$ ,

$\implies cos\theta = -\frac{29}{48}$ .

Using the identity $\cos2\theta = 2\cos^2\theta -1$ , we have that $\cos2\theta = -\frac{311}{1152}$ (a calculator could have been used here instead).

Knowing that the angle at the centre is twice the angle at the circumference, we can form a triangle of lengths $a, \sqrt{ \frac{252}{209}}$ and $\sqrt{ \frac{252}{209}}$ , where the angle opposite to $a$ is $2\theta$ .

By using the Cosine Rule a second time, we have that $a^2 = \frac{504}{209} - \frac{504}{209}\cos2\theta$ ,

$\implies a^2 = \frac{49}{16}$ (by substituting $\cos2\theta = -\frac{311}{1152}$ ),

$\implies a = \frac{7}{4}$

The longest side of the triangle is $a$ , which implies that $m = 7$ and $n = 4$ ,

$\therefore m^n = 7^4 = 2401$

The diagram below, in which x $= \theta$ , r = $\sqrt\frac{252}{209}$ , and C is the center of the circumcircle may also aid my explanation (or not!)

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Let the sides be $a=\dfrac{3}{2}x$ , $b=x$ and $c=\dfrac{2}{3}x$ . Now use the formula for the circumradius:

$R=\dfrac{abc}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}$

$6\sqrt{\dfrac{7}{209}}=\dfrac{x^3}{\dfrac{\sqrt{1463}}{36}x^2}$

$\dfrac{\sqrt{7}}{\sqrt{209}}=\dfrac{6x}{\sqrt{7} \sqrt{209}}$

$x=\dfrac{7}{6}$

Finally, the longest side is $a$ :

$a=\dfrac{3}{2} \cdot \dfrac{7}{6}=\dfrac{7}{4} \implies m=7 \implies n=4$

So, the final answer is $7^4=\boxed{2401}$ .