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1968 IMO Problems/Problem 1

Revision as of 00:44, 22 August 2024 by Pf02 (talk | contribs)

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 which could be the sides of a triangle.

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 could 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.


Solution 4

See Also

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