Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "==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== In tr...")
 
Line 17: Line 17:
  
 
Now assume that <math>b</math> is the greatest of the three consecutive integers. Then either <math>b|b-2</math> or <math>b|(b-2)^2</math>, depending on if <math>a=b-2</math> or <math>c=b-2</math>. <math>b|b-2</math> is absurd, so <math>b|(b-2)^2</math>, and <math>b|(b-2)^2-b^2+4b=4</math>. Therefore <math>b</math> is 1, 2, or 4. However, all of these cases are either degenerate or have been previously ruled out, so <math>b</math> cannot be the greatest of the three consecutive integers.
 
Now assume that <math>b</math> is the greatest of the three consecutive integers. Then either <math>b|b-2</math> or <math>b|(b-2)^2</math>, depending on if <math>a=b-2</math> or <math>c=b-2</math>. <math>b|b-2</math> is absurd, so <math>b|(b-2)^2</math>, and <math>b|(b-2)^2-b^2+4b=4</math>. Therefore <math>b</math> is 1, 2, or 4. However, all of these cases are either degenerate or have been previously ruled out, so <math>b</math> 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. <math>\blacksquare</math>
 
This shows that there is exactly one triangle with this property - and it has side lengths of 4, 5, and 6. <math>\blacksquare</math>
  
Line 23: Line 22:
  
 
{{alternate solutions}}
 
{{alternate solutions}}
 +
Alternate Solution:
 +
if in a triangle one angle is twice the other . Say in tr. ABC angle A=2angle B
 +
A=2B
 +
which implies C=180-3B
 +
SinC=Sin3B
 +
Sin^2A = Sin^2 2B= 2sinBcosBSin2B = sinB(SinB + Sin3B) = SinB(SinB + SinC)
 +
Hence, '''a^2 = b(b+c)'''
 +
using the above relation we check for triangle with consecutive sides.
 +
Putting b as the smallest,
 +
(b+2)^2 = b^2 +b(b+1)
 +
(b-4)(b+1)=0
 +
b=4,c=5,a=6
 +
putting similar cases we can show that all other solutions are non-integral
  
 
==See Also==
 
==See Also==

Revision as of 07:03, 9 January 2013

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

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. Alternate Solution: if in a triangle one angle is twice the other . Say in tr. ABC angle A=2angle B A=2B which implies C=180-3B SinC=Sin3B Sin^2A = Sin^2 2B= 2sinBcosBSin2B = sinB(SinB + Sin3B) = SinB(SinB + SinC) Hence, a^2 = b(b+c) using the above relation we check for triangle with consecutive sides. Putting b as the smallest, (b+2)^2 = b^2 +b(b+1) (b-4)(b+1)=0 b=4,c=5,a=6 putting similar cases we can show that all other solutions are non-integral

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1968_IMO_Problems/Problem_1&oldid=50478"