Pentagon and Circumcircle

Geometry Level pending

Let A A be the perimeter ratio of the outer pentagon to the inner pentagon. Let B B be the ratio of the inner pentagon's circumcircle to one of the five smaller internal circle's circumference. Then what can we say about the ratio A : B ? A:B?

Greater than two More than one, less than two Equal to 1 Less than one

W Rose by W Rose , 63, USA

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.

1 solution

Maharnab Mitra
Jan 26, 2016

D is the mid point of side AB (length a) and O is the circumcentre. So, radius of smaller circles is a 2 \frac{a}{2} .

Radius of circumcircle = a 2 sin ⁡ ( 3 6 o ) \frac{a}{2 \sin(36^o)}

Thus, B = 1 sin ⁡ ( 3 6 o ) \frac{1}{ \sin(36^o)}

You will find Δ C M A ≅ Δ O A D \Delta CMA \cong \Delta OAD . So, CM = NE = OD = a 2 tan ⁡ ( 3 6 o ) \frac{a}{2 \tan(36^o)}

Thus, CE = a + 2 × a 2 tan ⁡ ( 3 6 o ) a+ 2\times \frac{a}{2 \tan(36^o)} = a + a tan ⁡ ( 3 6 o ) =a+ \frac{a}{ \tan(36^o)}

On solving, A comes out to be 1 + 1 tan ⁡ ( 3 6 o ) 1+ \frac{1}{\tan(36^o)}

So, A:B comes out to be sin ⁡ ( 3 6 o ) + cos ⁡ ( 3 6 o ) = 1.396 \sin(36^o) + \cos(36^o) = 1.396

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Excellent presentation of your solution for this problem!

W Rose - 5 years, 4 months ago

Log in to reply

Thank you very much. :)

Maharnab Mitra - 5 years, 4 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...