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Geometry Level 5

As shown in the figure above, two quarter circles are inscribed in a square of side length 5. A smaller square is inscribed in between the two quarter circles and a circle of radius $r$ is inscribed in between the smaller square and the two quarter circles. What is $r$ ?

The answer is 0.6093.

3 solutions

Niranjan Khanderia
May 13, 2016

How could EF equald BF?

Axel Samyn - 5 years, 1 month ago

Thanks. It is an error. I am correcting it.

Niranjan Khanderia - 5 years, 1 month ago

In the last step, AF is 5-r

Michele Chiminazzo - 5 years, 1 month ago

Thanks. I have corrected.

Niranjan Khanderia - 5 years, 1 month ago

In the 2nd line, BF=5/2 (why?); and in the 4th, BF=5. How???

Nestor Taham - 5 years, 1 month ago

Thank you. BF=5/2 is an error. BF=5=radius of quarter circle. I am correcting.

Niranjan Khanderia - 5 years, 1 month ago

How do you know 5 - r =AF? You are asuming that A, F and the tangent point are colinear. They probably are, but how can you be 100% sure?

Cristian Juarez - 5 years, 1 month ago

Perpendicular to the tangent at the point of tangency is the diameter of the circle. So the centers are on this line. This is always so. Therefore when two circles are tangential we have, distance between two centers $d= r_1 \pm r_2.$ +for external contact, - for internal contact. I hope I am clear.

Niranjan Khanderia - 5 years, 1 month ago

Roger Erisman
May 13, 2016

Begin at lower left hand corner and label points (clockwise) A, B, C, D on large square. Start at lower rh corner of small square E, F, G, H. Center of circle is J

let h = side of small square.

Distance from D to G is 5, the radius of the arc. Distance from D to E is (1/2)*(5 - h).

Pythagorean's Theorem says : h^2 + (DF)^2 = 5^2 Since DF = 5 + (1/2)(5-h) we end up with h^2 + (1/4) h^2 + (5/2) h +25/4 = 25

Multiply through by 4

4 h^2 + h^2 +10 h + 25 = 100

5 h^2 + 10 h - 75 = 0

h^2 + 2*h - 15 = 0

(h+5)*(h-3) = 0

h = -5 or 3 discard negative h = 3

let r = radius of circle and let P = midpoint of EF

Triangle AJP is a right triangle with sides (3+r) and 2.5. Its hypotenuse is (5-r).

(3+ r)^2 + 2.5^2 = (5 - r) ^2

9 + 6 r + r^2 + 6.25 = 25 -10 r +r^2

16*r = 9.75

r = 0.6094

Vinod Kumar
Aug 22, 2020

solve

x^2+y^2=25,

(x-5/2)^2+(y-r-3)^2=r^2,

y=(r+3)(2/5)x,

x>0,y>0,r>0

r=39/64, x=800/281, y=1155/281

Answer r=39/64~0.609

Yeah! I reached 200,000 points

The answer is 0.6093.

3 solutions

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