Week 4 & S’mores
Last weekend, we built a small fire and had some S’mores in preparation for the camp this weekend in Chesterfield in the UK. Warmth and sweetness in winter are a must-have!
Anyway, let’s look at this week’s free problem!
The diagram shows eight circles of two different sizes. The circles are arranged in concentric pairs so that the centres form a square. Each larger circle touches one other larger circle and two smaller circles. The larger circles have radius 1. What is the radius of each smaller circle?
Give this problem a try before jumping in for the solution!
Solution
Let’s first annotate the diagram.
The centres of the circles are P, Q, R and S respectively.
The radius we want to compute is denoted r, it is the radius of each smaller circle.
We now construct a right-angled triangle with the centres P, Q and R.
Notice that each side of the right-angled triangle can be broken down as a sum of the larger radius and the smaller radius.
s = 1 + r
As for the hypotenuse, it’s simply the sum of two radii of the bigger circles.
h = 1 + 1 = 2
In summary, we have the following right-angled triangle.
To find the radius, we simply have to apply the Pythagoras’ Theorem.
Upon solving for r, we should get two solutions.
We reject the negative solution since it’s physically impossible to have a negative radius.
Therefore, we have the following as the radius.
The answer is C.
Well done.
What was your approach?
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Happy reading,
Barry 🍩