Difference between revisions of "1968 IMO Problems/Problem 1"

Latest revision as of 08:47, 9 March 2019

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. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Solution 2

[Incomplete, please edit] 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, $(*)$ $a^2 = b(b+c)$ (with assumptions. This needs clearing up) If $b$ is the shortest side, $(b+2)^2 = b^2 +b(b+1)$ $\implies (b-4)(b+1)=0$ , $\implies b=4, c=5, a=6$ , No other permutation of $a$ , $b$ and $c$ in terms of size gives integral values to $(*)$ [show]. So there is only one such triangle.

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.

@@ Line 3: / Line 3: @@
 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==
+==Solution 1==
 In triangle <math>ABC</math>, let <math>BC=a</math>, <math>AC=b</math>, <math>AB=c</math>, <math>\angle ABC=\alpha</math>, and <math>\angle BAC=2\alpha</math>. Using the [[Law of Sines]] gives that
@@ Line 22: / Line 22: @@
 {{alternate solutions}}
-Alternate Solution:
-if in a triangle one angle is twice the other . Say in tr. ABC angle A=2angle B
+==Solution 2==
-A=2B
+[Incomplete, please edit]
-which implies C=180-3B
+In a given triangle <math>ABC</math>, let <math>A=2B</math>, <math>\implies C=180-3B</math>, and <math>\sin C=\sin 3B</math>.
-SinC=Sin3B
+Then <cmath> \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)</cmath>
-Sin^2A = Sin^2 2B= 2sinBcosBSin2B = sinB(SinB + Sin3B) = SinB(SinB + SinC)
+Hence, <math>(*)</math> <math>a^2 = b(b+c)</math> (with assumptions. This needs clearing up)
-Hence, '''a^2 = b(b+c)'''
+If <math>b</math> is the shortest side, <math>(b+2)^2 = b^2 +b(b+1)</math>
-using the above relation we check for triangle with consecutive sides.
+<math>\implies (b-4)(b+1)=0</math>,
-Putting b as the smallest,
+<math>\implies b=4, c=5, a=6</math>,
-(b+2)^2 = b^2 +b(b+1)
+No other permutation of <math>a</math>, <math>b</math> and <math>c</math> in terms of size gives integral values to <math>(*)</math> [show]. So there is only one such triangle.
-(b-4)(b+1)=0
-b=4,c=5,a=6
+==Solution 3==
-putting similar cases we can show that all other solutions are non-integral
+NO TRIGONOMETRY!!!
+Let <math>a, b, c</math> be the side lengths of a triangle in which <math>\angle C = 2\angle B.</math>
+Extend <math>AC</math> to <math>D</math> such that <math>CD = BC = a.</math> Then <math>\angle CDB = \frac{\angle ACB}{2} = \angle ABC</math>, so <math>ABC</math> and <math>ADB</math> are similar by AA Similarity. Hence, <math>c^2 = b(a+b)</math>. Then proceed as in Solution 2, as only algebraic manipulations are left.
 ==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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