IMC 2020 Q23
23. Five congruent triangles, each of which is half a square, are placed together edge to edge in three different ways as shown to form shapes P, Q and R.
Which of the following lists gives the shapes in ascending order of the lengths of their perimeters?
- A P, Q, R
- B Q, P, R
- C R, Q, P
- D R, P, Q
- E P, R, Q
Intermediate Mathematical Challenge 2020, Question 23. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: A P, Q, R
Let the small sides of each triangle have length \(r\). This is also the side of the original square.
The longer side of each triangle is \(\sqrt{2}r\), since that is the diagonal of the square.
Hence the perimeters of shapes P, Q, R are \((2 + 3\sqrt{2})r\), \((6 + \sqrt{2})r\) and \((4 + 3\sqrt{2})r\) respectively.
Now \(6 + \sqrt{2} - (2 + 3\sqrt{2}) = 4 - 2\sqrt{2} = 2(2 - \sqrt{2})\), which is greater than zero.
Therefore the perimeter of P is less than the perimeter of Q.
Also, \(4 + 3\sqrt{2} - (6 + \sqrt{2}) = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)\), which is also greater than 0.
Therefore the perimeter of Q is less than the perimeter of R.
Hence, in ascending order, the lengths of the perimeters are P, Q, R.
Solution © UK Mathematics Trust, from the IMC 2020 solutions. Longer versions with investigations: Solutions and investigations (PDF).