Hey, this is Presh Talwalkar. Here's an interesting question. We have a large circle with chord PQ. Above the chord, we have a small circle that's tangent to the chord and touching the large circle. Below the chord PQ, we have another circle that's tangent to the chord and touching the large circle, and the two small circles are touching each other.
All three circle centers are aligned. If PQ is equal to 6, What is the area that's shaded in blue, which is the area between the two small circles and the large circle? This is from the UKMT IMC 2023 question 25 and I thank Max and Asir for the suggestion. Pause the video if you'd like to give this problem a try and when you're ready keep watching to learn how to solve this problem. I'll present two ways to solve the problem.
The first method uses a right triangle, and the second method uses the chord-chord power theorem. Let's get started with method 1. Let's work this out carefully. Suppose the top circle has radius equal to a, suppose the bottom circle has radius equal to b, and suppose that b is greater than a.
Now the diameter of the top circle will be equal to 2a, and the diameter of the bottom circle will be equal to 2b. Suppose the large circle has radius equal to r. We know that the span of the two diameters of the small circles will span the diameter of the large circle.
So we have 2r is equal to 2a plus 2b and that means r is equal to a plus b. Now we want to calculate the shaded area. This will be the area of the large circle minus the areas of the small circles. So it will be pi r squared minus pi a squared minus pi b squared. But we know that r is equal to a plus b, so we can substitute in.
Then we can expand out this binomial. Now we can do some cancellation. We can cancel out pi a squared and pi b squared, leaving the shaded area is equal to 2ab pi. So this is our goal that we want to calculate and we'll keep that in mind.
And we know the radius of the large circle is equal to a plus b. Let's find the distance from the center of this circle to the chord PQ. How can we do that? We'll draw this radius of the large circle and we know that it's equal to a plus b.
Then we want to subtract out the diameter of the small circle. So we want to do a plus b minus 2a. So this will be r minus 2a and that'll be a plus b minus 2a which is equal to b minus a.
So this distance is equal to b minus a. Now Let's consider this right triangle where one leg is b minus a, the hypotenuse is a plus b. What's the other leg equal to? It's exactly half of pq, so this triangle will have one leg that's equal to 3, which is one half of 6. So now in this right triangle, we have the square of a plus b is equal to the square of b minus a plus the square of 3. We can expand the binomials then cancel out the a squared and b squared terms, so we get that 4ab is equal to 3 squared. But we know we want to calculate 2ab pi, so we divide this equation by 2, and then multiply both sides of the equation by pi, and that gives us the area 2ab pi is equal to 9 pi over 2. And that's one way to solve the problem.
Now on to method 2. We will use the chord-chord power theorem. So let's start the problem at this step where we've calculated the radius of the large circle is equal to a plus b and we want to calculate the shaded area which is 2ab pi. So let's say the diameter of the large circle has endpoints of s and t and it intersects the chord pq at m. By the chord-chord power theorem we have sm multiplied by mt is equal to pm multiplied by mq. But SM is equal to 2A, MT is equal to 2B, and then PM is equal to MQ, which is equal to half of PQ, which equals 3. So we have 2A multiplied by 2B is equal to 3 times 3, which means 4AB is equal to 9. We now divide both sides by 2 and multiply by pi to get that 2AB pi is equal to 9 pi over 2. And that's another way to solve the problem.
Thanks for making us one of the best communities on YouTube. See you next episode of Mind Your Decisions, where we solve the world's problems, one video at a time.