Problem of the Week

Problem 26 Solution

Congratulations to Markus Kuan from Northstowe Secondary College, Matthew Whitten from The Cavendish School, Anna Poliakoff from Hitchin Girls’ School, Imogen Ficken from Birchwood High School and Riddhiman Kumar from Cambourne Village College for successfully solving the problem!

Consider triangle ABC:

Horizontal length: $4 - (-1) = 5$

Perpendicular height: $2 - 0 = 2$

Area of triangle ABC: $\frac{5 \times 2}2 = 5$

Therefore, each area will constitute 2.5 square units.

Let the point intersecting $y=mx$ and AC as D.

$OC = 4$ , area of OCD is required to be 2.5.

Therefore, we can work out the y-coordinate of D:

Horizontal distance = $OC = 4$
$4(y-0)/2 = 2.5$
$y = 1.25$

Work out Equation AC:
y-intercept = 2

Plug C(4,0) into $y=mx+2$ :
$0=4m+2$
$m=-\frac{1}2$
Equation AC: $y=-\frac{1}2 x + 2$

Point D is on AC, plug D’s y-coordinate into Equation AC to find it’s x-coordinate:

$1.25=-\frac{1}2 x + 2$

$-0.75=-\frac{1}2 x$

$1.5=x$

D(1.5,1.25), Substitute into $y=mx$ :

$1.25 = 1.5m$

$m = \frac{5}6$