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Problem 1209

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Algebra Difficulty 5.2 Prove it The South African Mathematical Olympiad Third Round · South Africa

Given that
$\frac{a-b}{c-d} = 2 \quad \text{and} \quad \frac{a-c}{b-d} = 3$
for certain real numbers $a$ , $b$ , $c$ , $d$ , determine the value of
$\frac{a-d}{b-c}$

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solutions — 2

Solution 1

Set $x = c - d$ and $y = b - d$ . We have
$a - d = (c - d) + (a - c) = x + 3y$
and
$a - d = (b - d) + (a - b) = y + 2x,$
hence $x + 3y = y + 2x$ , which implies $x = 2y$ . Now we get $a - d = 5y$ and
$b - c = (b - d) - (c - d) = y - x = -y,$
so

$\frac{a-d}{b-c}$ = -5.

Solution 2

We are given that
$a - b = 2c - 2d, \text{ thus } a + 2d = b + 2c,$
and
$a - c = 3b - 3d, \text{ thus } a + 3d = 3b + c.$
Multiply the first equation by 4 and the second one by 3, and subtract:
$a - d = 4(a + 2d) - 3(a + 3d) = 4(b + 2c) - 3(3b + c) = -5b + 5c,$
thus

$\frac{a-d}{b-c}$ = -5.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.