1968 IMO Problems/Problem 1

Problem

Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another.

Solution 1

In triangle $ABC$ , let $BC=a$ , $AC=b$ , $AB=c$ , $\angle ABC=\alpha$ , and $\angle BAC=2\alpha$ . Using the Law of Sines gives that

$\frac{b}{\sin{\alpha}}=\frac{a}{\sin{2\alpha}}\Rightarrow \frac{\sin{2\alpha}}{\sin{\alpha}}=2\cos{\alpha}=\frac{a}{b}$

Therefore $\cos{\alpha}=\frac{a}{2b}$ . Using the Law of Cosines gives that

$\cos{\alpha}=\frac{a^2+c^2-b^2}{2ac}=\frac{a}{2b}$

This can be simplified to $a^2c=b(a^2+c^2-b^2)$ . Since $a$ , $b$ , and $c$ are positive integers, $b|a^2c$ . Note that if $b$ is between $a$ and $c$ , then $b$ is relatively prime to $a$ and $c$ , and $b$ cannot possibly divide $a^2c$ . Therefore $b$ is either the least of the three consecutive integers or the greatest.

Assume that $b$ is the least of the three consecutive integers. Then either $b|b+2$ or $b|(b+2)^2$ , depending on if $a=b+2$ or $c=b+2$ . If $b|b+2$ , then $b$ is 1 or 2. $b$ couldn't be 1, for if it was then the triangle would be degenerate. If $b$ is 2, then $b(a^2+c^2-b^2)=42=a^2c$ , but $a$ and $c$ must be 3 and 4 in some order, which means that this triangle doesn't exist. therefore $b$ cannot divide $b+2$ , and so $b$ must divide $(b+2)^2$ . If $b|(b+2)^2$ then $b|(b+2)^2-b^2-4b=4$ , so $b$ is 1, 2, or 4. Clearly $b$ cannot be 1 or 2, so $b$ must be 4. Therefore $b(a^2+c^2-b^2)=180=a^2c$ . This shows that $a=6$ and $c=5$ , and the triangle has sides that measure 4, 5, and 6.

Now assume that $b$ is the greatest of the three consecutive integers. Then either $b|b-2$ or $b|(b-2)^2$ , depending on if $a=b-2$ or $c=b-2$ . $b|b-2$ is absurd, so $b|(b-2)^2$ , and $b|(b-2)^2-b^2+4b=4$ . Therefore $b$ is 1, 2, or 4. However, all of these cases are either degenerate or have been previously ruled out, so $b$ cannot be the greatest of the three consecutive integers. This shows that there is exactly one triangle with this property - and it has side lengths of 4, 5, and 6. $\blacksquare$

Solution 2

(Note: this proof is an expansion by pf02 of an outline of a solution posted here before.)

In a given triangle $ABC$ , let $A=2B$ , $\implies C=180-3B$ , and $\sin C=\sin 3B$ . Then

$\sin ^2 A = \sin ^2 2B = 2 \sin B \cos B \sin 2B = \sin B(\sin B + \sin 3B) = \sin B(\sin B + \sin C)$

Hence,

$a^2 = b(b + c)\ (*)$

Indeed, we know from the Law of Sines that

$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ .

Denote this ratio by $r$ ; we have $\sin A = ra, \sin B = rb, \sin C = rc$ . Substitute in $\sin ^2 A = \sin B(\sin B + \sin C)$ and simplify by $r^2$ . We get $(*)$ .

At this point, notice that $(*)$ is equivalent to the equality $a^2c = b(a^2 + c^2 - b^2)$ from Solution 1. Indeed, the latter can be rewritten as $a^2(c - b) = b(c + b)(c - b)$ , and we know that $c \ne b$ . So we could simply quote the fact (proven in Solution 1) that if $a, b, c$ are consecutive integers and $a^2 = b(c + b)$ , then $b = 4, c = 5, a = 6$ is the only solution.

For the sake of completeness, and for fun, I give a slightly different proof here.

We have six possibilities, depending on how the three consecutive numbers are ordered. The six possibilities are:

1: $a = b - 2, c = b - 1, b$

2: $c = b - 2, a = b - 1, b$

3: $a = b - 1, b, c = b + 1$

4: $c = b - 1, b, a = b + 1$

5: $b, a = b + 1, c = b + 2$

6: $b, c = b + 1, a = b + 2$

For each case, we substitute $a, c$ in $(*)$ , get an equation in $b$ , solve it, and get all the possible solutions. As a shortcut, notice that (*) implies that $b|a^2$ . If $a, b$ are consecutive integers, then they are relatively prime, so $b|a^2$ can not be true unless $b = 1$ . In this case the triangle would have sides $1, 2, 3$ , which is impossible. This eliminates cases 2, 3, 4 and 5.

In case 1, $(*)$ becomes

$(b - 2)^2 = b(b + b - 1)$ , or $b^2 + 3b - 4 = 0$ .

This has solutions $1, -4$ . The value $b = -4$ is impossible. The value $b = 1$ yields $a = -1, c = 0$ , which is impossible.

In case 6, $(*)$ becomes

$(b + 2)^2 = b(b + b + 1)$ , or $b^2 - 3b - 4 = 0$ .

The solutions are $-1, 4$ . The value $b = -1$ is impossible. Thus, we get the unique triangle $a = 6, b = 4, c = 5$ .

Solution 3

NO TRIGONOMETRY!!!

Let $a, b, c$ be the side lengths of a triangle in which $\angle C = 2\angle B.$

Extend $AC$ to $D$ such that $CD = BC = a.$ Then $\angle CDB = \frac{\angle ACB}{2} = \angle ABC$ , so $ABC$ and $ADB$ are similar by AA Similarity. Hence, $c^2 = b(a+b)$ . Then proceed as in Solution 2, as only algebraic manipulations are left.

1968 IMO (Problems) • Resources
Preceded by First Problem	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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1968 IMO Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4

See Also