IMC 2020 Q18
18. The diagram shows an isosceles right-angled triangle which has a hypotenuse of length \(y\). The interior of the triangle is split up into identical squares and congruent isosceles right-angled triangles.
What is the total shaded area inside the triangle?
- A \(\dfrac{y^2}{2}\)
- B \(\dfrac{y^2}{4}\)
- C \(\dfrac{y^2}{8}\)
- D \(\dfrac{y^2}{16}\)
- E \(\dfrac{y^2}{32}\)
Intermediate Mathematical Challenge 2020, Question 18. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C \(\dfrac{y^2}{8}\)
Let the length of each equal side of the given triangle be \(x\).
Then, by Pythagoras’ theorem, \(x^2 + x^2 = y^2\). So \(x^2 = \tfrac{y^2}{2}\).
In the triangle, four of the squares are shaded while the unshaded area consists of two squares and four half-squares.
Therefore, half of the area of the triangle is shaded.
Now the area of the triangle is \(\tfrac{1}{2} \times x \times x = \tfrac{1}{2}x^2\).
Therefore the total shaded area of the triangle is \(\tfrac{1}{4}x^2 = \tfrac{1}{4} \times \tfrac{y^2}{2} = \tfrac{y^2}{8}\).
Solution © UK Mathematics Trust, from the IMC 2020 solutions. Longer versions with investigations: Solutions and investigations (PDF).