IMC 2015 Q21
21. A rectangle is placed obliquely on top of an identical rectangle, as shown.
The area \(X\) of the overlapping region (shaded more darkly) is one eighth of the total shaded area.
What fraction of the area of one rectangle is \(X\)?
- A \(\dfrac{1}{3}\)
- B \(\dfrac{2}{7}\)
- C \(\dfrac{1}{4}\)
- D \(\dfrac{2}{9}\)
- E \(\dfrac{1}{5}\)
Intermediate Mathematical Challenge 2015, Question 21. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D \(\dfrac{2}{9}\)
Let the area of each rectangle be \(Y\). Then the total shaded area is \(2(Y - X) + X = 2Y - X\). Therefore \(X = \tfrac{1}{8}(2Y - X)\). So \(8X = 2Y - X\), that is \(9X = 2Y\). Therefore \(\tfrac{X}{Y} = \tfrac{2}{9}\).
Solution © UK Mathematics Trust, from the IMC 2015 solutions. Longer versions with investigations: Solutions and investigations (PDF).