# Enigmatic Code

Programming Enigma Puzzles

## Enigma 1250: Sheng-phooey

From New Scientist #2406, 2nd August 2003 [link]

A friend has hired a style consultant to help convert a few cowsheds into luxury homes. One the day we visited, the guru was in one of the rooms explaining how he had found the most harmonious position for the ceiling pendant light. He had drawn the rectangular ceiling to scale on a sheet of paper and cut it out. He had then taken the rectangle and made two mutually perpendicular cuts as shown in the figure: the resulting three separate shapes could then be rearranged to form a square of side representing 10 feet.

The shortest distance of point P on the ceiling from the nearest wall was 2 feet. This point was, we were told, the only possible position of the light fitting. I thought to mention the other equally valid positions, but it wasn’t my money buying this “expertise”.

What were the dimensions (length first) of the room?

[enigma1250]

### One response to “Enigma 1250: Sheng-phooey”

1. Jim Randell 8 March 2015 at 7:57 am

This puzzle is similar to Enigma 1758 (also the rectangle is the same as my first solution to Enigma 1546).

AC = BD = x, AB = CD = y, xy = 100, CR = PD = 10, QP = 2, PR = z.

From the similar triangles ACR, CDP, PQR in the diagram we have:

cos(θ) = x/10 = 10/y = 2/z

Hence:

y = 5z

Applying Pythagoras’ Theorem to CDP:

10² + (10 – z)² = y²

Substituting for y = 5z:

200 – 20z + z² = 25z²
24z² + 20z – 200 = 0
4(2z – 5)(3z + 10) = 0

Hence:

z = 5/2, y = 25/2, x = 8.

Solution: The dimensions of the room are 12′ 6″ × 8′.

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