Algebra and Geometry Don't Fit Well!

Geometry Level 5

$\large \dfrac {2x}{\sqrt 3} = y \sqrt 2 = \dfrac {2z}{\sqrt {2-\sqrt 3}}$

Consider a triangle with side lengths $x$ , $y$ and $z$ satisfying the equation above. If the longest side has length 303, then the circumradius of this triangle can be expressed as $a \sqrt b$ , where $a$ and $b$ are positive integers with $b$ square-free. Find $a+b$ .

The answer is 104.

2 solutions

Chew-Seong Cheong
Jul 24, 2016

$\begin{aligned} \frac {2x}{\sqrt 3} & = y \sqrt 2 = \frac {2z}{\sqrt{2-\sqrt{3}}} \\ \frac x{\frac {\sqrt 3}2} & = \frac y{\frac 1{\sqrt 2}} = \frac z {\frac 12 \sqrt{2-\sqrt{3}}} & \small \color{#3D99F6}{\text{Since }x, y, z \text{ are side lengths, the equation is the sine rule.}} \\ \frac x{\sin 120^\circ} & = \frac y{\sin 45^\circ} = \frac z{\sin 15^\circ} \end{aligned}$

The largest angle is $120^\circ$ , therefore, the longest side is $x=303$ .

The circumradius is given by $R = \dfrac x{2 \sin 120^\circ} = \dfrac {303}{\sqrt 3} = 101 \sqrt 3$ .

$\implies a+b = 101 + 3 = \boxed{104}$

Nice solution. Up voted.
But I have a problem. Can any one explain ? Thanks. It is as follows.
$\dfrac 2 {\sqrt6}=.816496<1, \ \ \ \therefore\ \ x>y\ and\ y=.816496x.\\ \sqrt{2- \sqrt3}=.517638<1,\ \ \ \therefore\ \ x>z\ and\ z=.517638x.\\ \therefore\ x=303.\\$ $R=\dfrac{x^3*y*z}{x^2\sqrt{(1+y+z)*(-1+y+z)*(1-y+z)*(1+y-z)} }\\$ $= \dfrac{303*.816496*.517638}{\sqrt{(1+.816496+.517638)*(-1+.816496+.517638)*(1-.816496+.517638)*(1+.816496-.517638)} } \\$ $=\sqrt{23090.40135,,,,,,,,}\\ Since \ R=a\sqrt b\ ,\text{ with a,b as integers and b square free,}\\ R^2=a^2*b\ \text{must\ be\ an\ integer.}\\$ $However\ with\ a\ \ 50\ \ digit\ calculator, \ R^2=23090.40135,,,,,,,,\\ \text{So the answer can not be in the form } \ a\sqrt b\ both\ integers.$

Niranjan Khanderia - 4 years, 10 months ago

$R^2 = \dfrac {x^6y^2z^2}{x^4(1+y+z)(-1+y+z)(1-y+z)(1+y-z)}$ . We note that $y^2z^2 = \dfrac 46 (2-\sqrt{3})$ is not rational. $(1+y+z)(-1+y+z)(1-y+z)(1+y-z) = y^4 + z^4 - 2y^2z^2 + 2y^2 + 2z^2$ . Note that $z^2$ and $z^4$ are not rational. Therefore, for $R=a\sqrt b$ , $a$ and $b$ are not rational.

Chew-Seong Cheong - 4 years, 10 months ago

Thanks very much for the explaination.

Niranjan Khanderia - 4 years, 10 months ago

I understood till x=303 but how did you find the circumradius? Is there any formula?

Satwik Murarka - 4 years, 10 months ago

Circumradius is given by: $R = \dfrac a{2\sin A}$ which was used above. Check out the wiki here .

Chew-Seong Cheong - 4 years, 10 months ago

Thank you I now understood it.

Satwik Murarka - 4 years, 10 months ago

Aaghaz Mahajan
May 17, 2018

A simple use of SINE RULE!!!! And look out for the value of SIN(15).......!!

Algebra and Geometry Don't Fit Well!

The answer is 104.

2 solutions

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