VERY DIFFICULT | UKMT Intermediate Maths Challenge 2023 last question

In this video, we tackle the final question from the 2023 UKMT Intermediate Maths Challenge, and it's a real brain-teaser! We are presented with a large circle containing two smaller circles inscribed within it, and a chord of length six. Our task is to find the shaded area, which is the region not occupied by the smaller circles. What makes this question fascinating is the clever approach we use to solve it, employing key concepts related to circles.

First, we establish the radii of the two smaller circles, denoting them as R1 and R2. By recognizing that the smaller circles lie on a diameter of the larger circle, we can determine that the diameter of the larger circle is 2R2 + 2R1. To find the radius of the larger circle, we divide this diameter by 2, giving us R1 + R2.

By exploring the areas of the larger circle and the two smaller circles, we arrive at a simple expression for the shaded area: 2πR1R2. However, we can't stop there. We still have a valuable piece of information—the chord length of six—to incorporate into our solution.

To do this, we introduce a new radius, which forms a right-angled triangle inside the larger circle. We label the missing side of this triangle as R1 - R2, which represents the height of the triangle. By applying Pythagoras's Theorem to this triangle, we derive a relationship between R1, R2, and the chord length.

Simplifying the equation, we find that R1R2 = 9/4. Finally, substituting this result into our formula for the shaded area, we obtain a numerical value: 9π/2.

Join us in this exciting journey as we navigate through the intricacies of circle geometry and unlock the beautiful answer to this very difficult question! Don't forget to like, share, and subscribe for more math challenges and fascinating solutions. Let's dive in!

[mancubsofficial]

Wow what a question! Sometimes when doing these questions, you don't know how to proceed because you forget that you can draw extra lines to help you! Once you showed the right-angled triangle with that other radius, it suddenly made more sense! Great video! :)

[nithingeorge]

This is how I solved it:
Required Area: pi*(r1+r2)^2 - pi*r1^2 - pi*r2^2 = 2*pi*r1*r2
Call the diameter AB, A on the top end and B on the bottom end. Call the point of contact of the 2 inner circles C.
Angle PBC = Angle ABC = x.
PC = 3
tan x = 2*r1/3 = 3/(2*r1)
=> 4*r1*r2 = 9
Required Area = 2*pi*r1*r2 = (9*pi)/2