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

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==Solution 2==
 
==Solution 2==
 
[Incomplete, please edit]
 
[Incomplete, please edit]
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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>
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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>.
 
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>
 
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>
 
Hence, <math>(*)</math> <math>a^2 = b(b+c)</math> (with assumptions. This needs clearing up)
 
Hence, <math>(*)</math> <math>a^2 = b(b+c)</math> (with assumptions. This needs clearing up)
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<math>\implies (b-4)(b+1)=0</math>,
 
<math>\implies (b-4)(b+1)=0</math>,
 
<math>\implies b=4, c=5, a=6</math>,
 
<math>\implies b=4, c=5, a=6</math>,
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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.<cmath>\blacksquare</cmath>.
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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.
  
 
==Solution 3==
 
==Solution 3==
 
NO TRIGONOMETRY!!!
 
NO TRIGONOMETRY!!!
  
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Let <math>a, b, c</math> be the side lengths of a triangle in which <math><C = 2<B.</math>
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Let <math>a, b, c</math> be the side lengths of a triangle in which <math>\angle C = 2\angle B.</math>
  
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Extend <math>AC</math> to <math>D</math> such that <math>CD = BC = a.</math> Then <math><CDB = <ACB/2 = <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.
+
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==
 
==See Also==

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.

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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