The Enclosed one

Geometry Level 4

The figure above shows two circles , each of radius 1 cm 1\text{ cm} passes through each center. A square is inscribed inside the region enclosed between these two circles. Find the side length of this square (in cm \text{cm} ).

Give your answer to 3 decimal places.


The answer is 0.822.

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2 solutions

Ahmad Saad
Jul 1, 2016

Nice one! But the Diagram needs a little correction...

nibedan mukherjee - 4 years, 11 months ago

What is the error in diagram?

Ahmad Saad - 4 years, 11 months ago

circles centers are not in the correct position.. ha ha! ;( just a joke.. but nice solution using CG .cheers!

nibedan mukherjee - 4 years, 11 months ago

yes, you are right. I'll correct it.

Ahmad Saad - 4 years, 11 months ago

Now its perfect... Upvoted!

nibedan mukherjee - 4 years, 11 months ago

Square with EB as a diagonal is quarter of square ABCD.
Also F is mid point of BC, E of GF and OO'. And BF=EF= say x. OE=EO'= 1 2 \frac12 .
Using Pythagoras Theorem in right triangle OBF.
O B 2 = O F 2 + B F 2 , 1 2 = ( 1 2 + x ) 2 + x 2. S o l v i n g q u a d r a t i c i n x , x = 0.4114. B u t B C = 2 x = 0.8228. OB^2=OF^2+BF^2, \\ \implies\ 1^2=(\frac12+x)^2 + x*2.\\ Solving \ quadratic\ in\ x, x=0.4114.\\ But\ BC=2x=0.8228.


Niranjan Khanderia - 4 years, 11 months ago

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Your approach is same as of mine.. only the difference is that you have used the upper portion of the square.. good solution though.. cheers!

nibedan mukherjee - 4 years, 11 months ago

nyc approach !

Alex Baron - 4 years, 11 months ago

If you want the exact answer without using graphs:

Take the side of the square to be 'a'

BE² = EF² + FB²

BE = √(2*(a/(2))²) = a/(√2)

Triangle BEO has sides BE = a/(√2), B0 = 1 and EO = 0.5

Angle BEO = 90° + 45° = 135°

Using cosine rule on triangle BEO,

1² = 0.5² + (a/(√2))² - 2 * 0.5 * a/(√2) * cos(135)

1 = 1/4 + a²/2 - a/(√2) * - √(2)/ 2

1 = 1/4 + a²/2 + a/2

After some rearranging...

2a² + 2a - 3 = 0

a = (√(7)-1)/2 (= 0.8228...)

Oliver G - 1 year, 7 months ago

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