Shaded Area Between Circles And A Chord – Mind Your Decisions

Thanks to Max W. and Asir for the suggestion! This problem is from the UKMT IMC 2023, Q25.

A circle has chord PQ. Above the chord (and below it) is a circle tangent to the chord and the large circle. The two inscribed circles are also tangent to each other. The three circle’s centers are collinear. If PQ = 6, what is the area between the two smaller circles and the large circle, shaded in blue?

As usual, watch the video for a solution.

Seemingly impossible area calculation

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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 Shaded Area Between Circles And A Chord

(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).

Thanks David H. for alerting me of typos!

This is an amazing problem because there is just enough information to work it out. Let the top circle have radius equal to a, the bottom b, and the large circle r. Let the large circle have center O and let M be the midpoint of PQ, so PM = MQ = 3.

Since the three circles centers are collinear, the diameter of the large circle is the sum of the diameters of the small circles. Thus we have:

2r = 2a + 2b
r = a + b

We wish to calculate the shaded area which is the area of the large circle minus the areas of the smaller circles. So we have:

πr² – πa² – πb²
= π(a + b)² – πa² – πb²
= π(a² + b² + 2ab) – πa² – πb²
= 2abπ

Now let’s work out the problem in two different ways.

Method 1: right triangle

Since the bottom circle is tangent to the chord PQ, its diameter is perpendicular to the chord PQ. Thus OM is perpendicular to PQ. The length of OM is the radius of the large circle minus the diameter of the small circle, so it is

OM
= r – 2a
= a + b – 2a
= b – a

We know have a right triangle OMQ with legs OM = b – a, MQ = 3, and hypotenuse OQ = r = a + b.

Thus we have:

(b + a)² = (b – a)² + 3²
b2 + a² + 2ab = b² + a² – 2ab + 9
2ab = – 2ab + 9
4ab = 9

But we need to calculate 2abπ, so multiplying both sides by π/2 gives:

2abπ = 9π/2

And that’s the answer!

Method 2: chord-chord power theorem

We can avoid calculating OM and having to work through the right triangle by using the chord-chord power theorem. Start from the step after calculating the radius of the large circle, and let ST be the diameter of the large circle.

By the chord-chord power theorem,

|SM||MT| = |PM||MQ|
(2a)(2b) = 3(3)
4ab = 9

Once again we can multiply both sides by π/2 to get the answer:

2abπ = 9π/2

Reference

The question UKMT IMC 2023 Paper, Q25
https://www.ukmt.org.uk/sites/default/files/ukmt/IMC_2023_Paper.pdf Q25

Solutions
https://www.ukmt.org.uk/sites/default/files/ukmt/IMC%202023%20Solutions.pdf

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