Four touching circles (Hula Hoops)

Geometry Level pending

You draw a square on the ground of side length 100 100 cm. Now you bring four Hula Hoops of diameter 80 80 cm, and you arrange them such that each hoop is tangent to a side of the square at its midpoint, and each pair of adjacent hoops are tangent to each other. How high above the ground (in cm) is the highest point of each of the hoops? If the answer is a b a \sqrt{b} for positive integers a a and b b , with b b square-free, then find a + b a + b


The answer is 45.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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

David Vreken
Sep 27, 2020

The hula hoops are the incircles of the lateral triangular faces of a pyramid with the square base of 100 cm 100 \text{ cm} that was drawn on the ground.

The base angle of one of the lateral faces is θ 1 = 2 tan ⁡ − 1 ( 1 2 ⋅ 80 1 2 ⋅ 100 ) = 2 tan ⁡ − 1 ( 4 5 ) \theta_1 = 2 \tan^{-1}(\frac{\frac{1}{2} \cdot 80}{\frac{1}{2} \cdot 100}) = 2 \tan^{-1} (\frac{4}{5}) , which makes the slant height 100 2 tan ⁡ θ 1 = 50 tan ⁡ ( 2 tan ⁡ − 1 ( 4 5 ) ) = 2000 9 cm \frac{100}{2} \tan \theta _1 = 50 \tan (2 \tan^{-1} (\frac{4}{5})) = \frac{2000}{9} \text{ cm} .

The dihedral angle between a lateral face and the base is then θ 2 = cos ⁡ − 1 ( 100 2 2000 9 ) = cos ⁡ − 1 ( 9 40 ) \theta_2 = \cos^{-1} (\frac{\frac{100}{2}}{\frac{2000}{9}}) = \cos^{-1} (\frac{9}{40}) , so the highest point of each hoop is 80 sin ⁡ θ 2 = 80 sin ⁡ ( cos ⁡ − 1 ( 9 40 ) ) = 14 31 cm 80 \sin \theta_2 = 80 \sin (\cos^{-1} (\frac{9}{40})) = 14\sqrt{31} \text{ cm} .

Therefore, a = 14 a = 14 , b = 31 b = 31 , and a + b = 14 + 31 = 45 a + b = 14 + 31 = \boxed{45} .

0 Helpful 0 Interesting 2 Brilliant 0 Confused

Excuse me, there’s one point I didn’t get. Could you please elaborate on why an isosceles triangle with base 100 and an incircle with diameter 80 has base angle 2 tan ⁡ − 1 ( 1 2 ⋅ 80 1 2 ⋅ 100 ) 2\tan ^{-1}(\frac{\frac12\cdot 80}{\frac12\cdot 100}) ? Thank you!

Jeff Giff - 8 months, 1 week ago

Log in to reply

The center of an incircle is also on the bisector of each angle, so from the right angle triangle formed by the radius 1 2 ⋅ 80 \frac{1}{2} \cdot 80 , half the base side 1 2 ⋅ 100 \frac{1}{2} \cdot 100 , and the bisector of the base angle, half the base angle would be tan ⁡ − 1 ( 1 2 ⋅ 80 1 2 ⋅ 100 ) \tan^{-1}(\frac{\frac{1}{2} \cdot 80}{\frac{1}{2} \cdot 100}) , which makes the whole base angle 2 tan ⁡ − 1 ( 1 2 ⋅ 80 1 2 ⋅ 100 ) 2 \tan^{-1}(\frac{\frac{1}{2} \cdot 80}{\frac{1}{2} \cdot 100}) .

David Vreken - 8 months, 1 week ago

Log in to reply

Oic! Thanks :)

Jeff Giff - 8 months, 1 week ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...