How big is the orange?

Geometry Level 3

A circle of radius x is inscribed in a square. What is the radius r r of the largest circle that can fit into one of the corners?

State the radius as r = x ( a + b c ) r = x(a + b \sqrt{c}) where a , b , c , a, b, c, are integers and c c is not divisible by the square of any prime. Find a + b + c a+b+c .


The answer is 3.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Guiseppi Butel
May 30, 2014

title title <br>a+b+c= 3+(-2)+2 = 3

I'm disappointed that there weren't more attempts at this problem. Its solution is comparatively easy. I was leading up to a similar case but in 3 dimensions which would really challenge you.

Guiseppi Butel - 7 years ago

Log in to reply

Have you posted the 3-dimensions case? If so, you should link it.

Calvin Lin Staff - 6 years, 5 months ago

Looking 4wrd to d 3D problem.By the way how about generalizing this problem for the nth corner circle.Can it be done?

Chandrachur Banerjee - 7 years ago

I did the same way..

Mark Vincent Mamigo - 6 years, 5 months ago

a = 3 , b = 1 , c = 8 a=3,b=-1,c=8 is also a valid answer, so maybe you should specify b , c b,c as coprime.

Jared Low - 6 years, 5 months ago
Esrael Santillan
Jul 22, 2014

Equating the 2 formulas for half of the diagonal of the square: x 2 + x 2 = x + r + r 2 + r 2 x 2 = x + r + r 2 x ( 2 1 ) = r ( 1 + 2 ) r = x ( 2 1 ) ( 2 + 1 ) r = x ( 2 1 ) ( 2 + 1 ) ( 2 1 ) ( 2 1 ) r = x 2 2 2 + 1 2 1 r = x ( 3 2 2 ) r = x ( a + b c ) , a = 3 , b = 2 , c = 2 \begin{aligned} \sqrt{x^2 + x^2} &= x + r + \sqrt{r^2 + r^2} \\ x\sqrt{2} &= x + r + r\sqrt{2} \\ x\left(\sqrt{2} - 1\right)&= r\left(1 + \sqrt{2}\right) \\ r &=x\frac{\left(\sqrt{2} - 1\right)}{\left(\sqrt{2} + 1\right)} \\ r &=x\frac{\left(\sqrt{2} - 1\right)}{\left(\sqrt{2} + 1\right)}\frac{\left(\sqrt{2} - 1\right)}{\left(\sqrt{2} - 1\right)} \\ r &= x\frac{2 - 2\sqrt{2}+1}{2-1} \\ r &= x\left( 3 -2\sqrt{2}\right) \\ r &= x\left( a + b\sqrt{c}\right), a = 3, b = -2, c = 2 \\ \end{aligned}

Thus, a + b + c = 3 2 + 2 = 3 a+b+c=3-2+2=\boxed{3}

completely wrong concept

nibedan mukherjee - 6 years, 10 months ago

Log in to reply

Can you elaborate, in what sense is this a "completely wrong concept"?

Calvin Lin Staff - 6 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...