UKMT Maclaurin 2023 - Q1

Math
Olympiad
UKMT
UKMT Maclaurin

Published on Sun Mar 19 2023. Views

My solution to the problem.

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UKMT Maclaurin 2023 Question 3 Solution

Problem Statement

A plank of wood has one end, A, against a vertical wall. Its other end, B, is on horizontal ground. When end A slips down 8cm, end B moves 4cm further away from the wall. When end A slips down a further 9cm, end B moves a further 3cm away from the wall. Find the length of the plank.

My Solution

Based on the given information, we can construct a diagram similar to the one below (NOT drawn to scale):

where:

$AA' = 8$
$A'A'' = 9$
$BB' = 4$
$B'B'' = 3$

Let

$AT = x$
$BT = y$
$t = AB = A'B' = A''B''$

Triangles $ABT$ , $A'B'T$ , and $A''B''T$ all have the same hypotenuse since the length of the wood plank remains unchanged as it slips.

Thus a system of equations can be formed:

\begin{aligned} \begin{cases} (x + 17)^2 + y^2 = t^2 \\ (x + 9)^2 + (y + 4)^2 = t^2 \\ x^2 + (y + 7)^2 = t^2 \\ \end{cases}\, \end{aligned}

Combining the first equation with the second:

\begin{aligned} (x + 17)^2 + y^2 &= (x + 9)^2 + (y + 4)^2 \\ x^2 + 34x + 289 + y^2 &= x^2 + 18x + 81 + y^2 + 8y + 16 \\ 16x + 192 &= 8y \\ 2x + 24 &= y\\ \end{aligned}

Combining the third equation with the second:

\begin{aligned} x^2 + (y + 7)^2 &= (x + 9)^2 + (y + 4)^2 \\ x^2 + y^2 + 14y + 49 &= x^2 + 18x + 81 + y^2 + 8y + 16 \\ 6y = 18x + 48 \\ y = 3x + 8 \end{aligned}

This obtains a new system of equations:

\begin{aligned} \begin{cases} y = 3x + 8 \\ y = 2x + 24 \end{cases} \end{aligned}

This gives $x = 16$ and $y = 56$ . This pair of values can be substituted into any of the three original equations to find $t$ and they will give the same value. However, I personally find the second one giving nicer numbers:

\begin{aligned} t^2 &= (x + 9)^2 + (y + 4)^2 \\ &= 25^2 + 60^2 = 625 + 3600 = 4225 \\ t &= \sqrt{4225} = 65 \end{aligned}