Wednesday, October 12, 2005

Interesting Maths Question

I saw this interesting maths question on TV.

PQRS is a rectangle with PQ = 15 units, QR = 6 units and it is divided into 4 smaller rectangles by two lines, TU and VW. T is a point on PS, U is a point on QR, V is a point on PQ and W is a point on RS. X is the intersection point between TU and VW. If the area of triangle SXW is 4 sq units and area of triangle XVQ is 16 sq units, find the area of rectangle XURW.

Here is my solution:

Let WR = x and RU = y.

This means that
(XURW) = xy
SW = (15 - x)
UQ = (6 - y)

(SXW) = 4
==> (TXWS) = 8
==> (15 - x)y = 8 ---- (1)

(XVQ) = 16
==> (VQUX) = 32
==> x(6 - y) = 32 ---- (2)

From (1), y = 8 / (15 - x) ---- (3)
Sub (3) into (2)
==> x(6 - 8 / (15 - x)) = 32
==> 6x (15 - x) - 8x = 32 (15 - x)
==> x^2 - 19x + 80 = 0
==> x = (19 +- sqrt(41)) / 2 ---- (4)

Sub (4) into (3)
==> y = (11 +- sqrt(41)) / 5 --- (5)

(4)*(5) ==> xy = 25 +- 3* sqrt(41)
==> (XURW) = 25 +- 3* sqrt(41) sq units

Both solutions for the area of rectangle XURW satisfy the consistency check that 0 < x < 15 and 0 < y < 6. Note that the assumption that S, X and Q are collinear is not true.

Check out a related link here

Afternote: By replacing the hardcoded values with variables using (TXWS) = A, (VQUX) = B, PQ = w, QR = h and k = wh - (A + B),

I found that (XURW) = (k +- sqrt(k^2 - 4AB)) / 2