IMC 2013 Q21
21. The square \(ABCD\) has an area of 196. It contains two overlapping squares; the larger of these squares has an area 4 times that of the smaller and the area of their overlap is 1. What is the total area of the shaded regions?
- A 44
- B 72
- C 80
- D 152
- E more information is needed
Intermediate Mathematical Challenge 2013, Question 21. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: B 72
The large square has area 196 = 142. So it has side-length 14. The ratio of the areas of the inner squares is 4 : 1, so the ratio of their side-lengths is 2 : 1. Let the side-length of the larger inner square be 2\(x\), so that of the smaller is \(x\). The figure is symmetric about the diagonal \(AC\) and so the overlap of the two inner squares is also a square which therefore has side-length 1. Thus the vertical height can be written as \(x + 2x - 1\). Hence \(3x - 1 = 14\) and so \(x = 5\). Also, the two shaded rectangles both have side-lengths \(2x - 1\) and \(x - 1\); that is 9 and 4. So the total shaded area is 72.
Solution © UK Mathematics Trust, from the IMC 2013 solutions. Longer versions with investigations: Solutions and investigations (PDF).