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Solution to puzzle 76 (Ladder against a box)

By similar triangles, y/9 = 9/x, and so xy = 81. (1)

By Pythagoras' Theorem, (x + 9)2 + (y + 9)2 = 402, or x2 + 18x + y2 + 18y + 162 = 1600. (2)

A 40 meter ladder resting against a building and upon a 9 meter cubed annex. Let the vertical distance from the top of the annex to where the ladder touches the wall be y. Let the horizontal distance of the bottom of the ladder from the annex be x.

Now note that (x + y)2 = x2 + 2xy + y2 = x2 + y2 + 162, from (1).

Hence (2) may be rewritten as: (x + y)2 + 18(x + y) − 1600 = 0.

Rejecting the negative root of this quadratic equation in x + y, we obtain x + y = 32.

Having found xy and x + y, we use Viète's formulas to write down the quadratic of which x and y are the roots: z2 − 32z + 81 = 0.

Assuming the ladder is inclined at more than 45° to the horizontal, so that x < y, we have x = 16 − 5root 7, y = 16 + 5root 7.

Therefore the height above the ground at which the ladder touches the building is 5(5 +root 7) meters.


Further reading

  1. [Java] The Longest Ladder
  2. [Java] Shortest Ladder over a Fence
  3. Crossed Ladders Problem

Source: Traditional

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