IMC 2008 Q25
25. A garden has the shape of a right-angled triangle with sides of length 30, 40 and 50. A straight fence goes from the corner with the right-angle to a point on the opposite side, dividing the garden into two sections which have the same perimeter. How long is the fence?
- A 25
- B \(8\sqrt{3}\)
- C \(5\sqrt{11}\)
- D \(5\sqrt{39}\)
- E \(12\sqrt{5}\)
Intermediate Mathematical Challenge 2008, Question 25. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: E \(12\sqrt{5}\)
In the diagram on the right, triangle \(ABC\) represents the garden, \(CD\) represents the fence and \(E\) is the foot of the perpendicular from \(D\) to \(AC\).
The two sections of the garden have the same perimeter so \(AD\) is 10 m longer than \(DB\). Hence \(AD\) = 30 m and \(DB\) = 20 m. As \(\angle AED\) and \(\angle ACB\) are both right angles, triangles \(AED\) and \(ACB\) are similar.
So \(\tfrac{AE}{AC} = \tfrac{AD}{AB} = \tfrac{30}{50}\). Hence \(AE = \tfrac{3}{5} \times 30\) m = 18 m. So \(EC\) = (30 − 18) m = 12 m.
Also, \(\tfrac{ED}{CB} = \tfrac{AD}{AB} = \tfrac{30}{50}\). Hence \(ED = \tfrac{3}{5} \times 40\) m = 24 m.
Finally, by Pythagoras' Theorem: \(CD^2 = EC^2 + ED^2 = (12^2 + 24^2)\) m2 = 5 × 122 m2.
So the length of the fence is \(12\sqrt{5}\) m.
Solution © UK Mathematics Trust, from the IMC 2008 solutions.