Home Puzzle 3 Problems About Finding The Length In A Given Figure – Mind Your Decisions

3 Problems About Finding The Length In A Given Figure – Mind Your Decisions

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3 Problems About Finding The Length In A Given Figure – Mind Your Decisions

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Problem 1

Thanks to Ismot from Bangladesh for the suggestion!

A right triangle ABC has legs AB = 4, BC = 6, and has a semicircle O with its center on the hypotenuse tangent to the legs AB at D and BC at E. What is the radius of the semicircle?

Problem 2

Thanks Michal from the Czech Republic for suggesting a problem that inspired this puzzle.

What is the value of x in the figure below?

Problem 3

This problem comes from the Bangladesh Mathematical Olympiad (BDMO) 2016.

Two large circles are tangent to each other and the parallel lines. Two small circles are tangent to the large circles and to one of the lines. If the distance between the centers of the small circles is 96, what is the distance between the centers of the large circles?

As usual, watch the video for a solution.

3 Problems About Finding The Length In A Given Figure

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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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Answer To Semicircle on Right Triangle Hypotenuse

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Problem 1

We will prove a more general formula. Let r be the radius of the semicircle. Then we have:

1/r = 1/AB + 1/BC
r = (AB)(BC)/(AB + BC)

The area of the entire right triangle is (1/2)(AB)(BC). Construct radii to the points of tangency.

The area of the entire triangle is also the sum of areas of the triangles AOB and BOC. Each has a height equal to r, and the base is a side of the right triangle. So we have:

area = (1/2)r(AB) + (1/2)r(BC)

Equating the two areas of ABC gives:

(1/2)(AB)(BC) = (1/2)r(AB) + (1/2)r(BC)
(AB)(BC) = r(AB + BC)
r = (AB)(BC)/(AB + BC)

Taking the reciprocal and simplifying gives:

1/r = 1/AB + 1/BC

The original question had legs of 4 and 6, so we have:

r
= 4(6)/(4 + 6)
= 12/5
= 2.4

Problem 2

Extend the slanted lines to construct rectangles and isosceles right triangles. The base consists of 2 hypotenuses of isosceles right triangles with legs of 6 and 4, so their lengths will be 6√2 and 4√2. The other part of the base is a leg of an isosceles triangle with a a leg equal to 10, so it is 10.

The total length is then equal to:

6√2 + 4√2 + 10
= 10√2 + 10
≈ 24.14

Problem 3

Circles that are tangent have their centers and the tangent point collinear. Let the large circles have a radius of a and the small circles b. Construct a right triangle between the center of a large circle, a small circle, and the tangent point of the two large circles. One leg will be a, the other leg will be ab, and the hypotenuse will be a + b. The distance between the two small circle centers is 2a – 2b, which is given to be 96.

From the diagram and given information we have:

2a – 2b = 96
a2 + (ab)2 = (a + b)2

From the first equation we have:

ab = 48
b = a – 48

We can substitute this into the second equation.

a2 + (ab)2 = (a + b)2
a2 + 482 = (a + a – 48)2
a2 + 482 = (2a – 48)2
a2 + 482 = 4a2 – 192a + 482
0 = 3a2 – 192a
0 = a(3a – 192)

Since a is non-zero, we have:

3a – 192 = 0
a = 64

We want the distance between the two large circles, and that is 2a = 128.

Reference

Problem 1 Math StackExchange
https://math.stackexchange.com/questions/2285180/prove-frac1r-frac1a-frac1b-for-a-semicircle-tangent-within

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