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Difference between revisions of "2006 Indonesia MO Problems/Problem 3"

(Solution to Problem 3 -- trig casework)
 
m (→‎Solution)
 
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'''Case 1: Two variables are 1s'''
 
'''Case 1: Two variables are 1s'''
  
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<br>
 
 
[[WLOG]], let <math>x,y = 1</math>.  Substitution results in <math>2+z=z</math>, resulting in no solutions.
 
[[WLOG]], let <math>x,y = 1</math>.  Substitution results in <math>2+z=z</math>, resulting in no solutions.
  
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'''Case 2: One variable is 1'''
 
'''Case 2: One variable is 1'''
  
−
<br>
 
 
[[WLOG]], let <math>x = 1</math>. Substitution results in <math>1+y+z=yz</math>, and rearranging the equation results in <math>yz-y-z=1</math>.  By [[Simon's Favorite Factoring Trick]], the equation can be rewritten as <math>(y-1)(z-1) = 2</math>.  Thus, <math>y=3</math> and <math>z=2</math> (or vice versa), so a solution is <math>(1,2,3)</math> in some permutation.
 
[[WLOG]], let <math>x = 1</math>. Substitution results in <math>1+y+z=yz</math>, and rearranging the equation results in <math>yz-y-z=1</math>.  By [[Simon's Favorite Factoring Trick]], the equation can be rewritten as <math>(y-1)(z-1) = 2</math>.  Thus, <math>y=3</math> and <math>z=2</math> (or vice versa), so a solution is <math>(1,2,3)</math> in some permutation.
  
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'''Case 3: No variables are 1s'''
 
'''Case 3: No variables are 1s'''
  
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<br>
 
 
Since we already found a solution permutation in Case 2, we want to show that there are no solutions in this case.  We can use [[induction]].
 
Since we already found a solution permutation in Case 2, we want to show that there are no solutions in this case.  We can use [[induction]].
  

Latest revision as of 11:00, 26 September 2019

Problem

Let $S$ be the set of all triangles $ABC$ which have property: $\tan A,\tan B,\tan C$ are positive integers. Prove that all triangles in $S$ are similar.

Solution

If all triangles in $S$ are similar, then the triplet of angles must be equivalent.


Let $\tan A = x, \tan B = y, \tan C = z$. Note that $A+B+C = 180^\circ$, so $z = \tan C = \tan (180^\circ - (A+B)) = -\tan(A+B)$.


By applying the Angle Addition Identity for Tangent, $-\tan(A+B) = -\frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{x+y}{xy - 1} = z$. Multiplying both sides by the denominator results in $x+y = xyz - z$, and rearranging results in $x+y+z = xyz$. We want this equation to be satisfied by positive integers, and we want to show that there is only one solution permutation. Thus, we can divide into cases based on the number of ones.


Case 1: Two variables are 1s

WLOG, let $x,y = 1$. Substitution results in $2+z=z$, resulting in no solutions.


Case 2: One variable is 1

WLOG, let $x = 1$. Substitution results in $1+y+z=yz$, and rearranging the equation results in $yz-y-z=1$. By Simon's Favorite Factoring Trick, the equation can be rewritten as $(y-1)(z-1) = 2$. Thus, $y=3$ and $z=2$ (or vice versa), so a solution is $(1,2,3)$ in some permutation.


Case 3: No variables are 1s

Since we already found a solution permutation in Case 2, we want to show that there are no solutions in this case. We can use induction.


For the base case, $2+2+2 < 2 \cdot 2 \cdot 2$. For the inductive step, assume $x+y+z < xyz$.


WLOG, let $z$ be the variable that is increased. Adding $xy$ to both sides results in $xy(z+1) > x+y+z + xy$. Since $x,y = 1$, $xy > 1$, so $x+y+z+xy > x+y+z+1$. Thus, $xy(z+1) > x+y+(z+1)$, so the inductive step is complete. Therefore, by induction, there are no solutions in this case.


Since all of the solutions are permutations of $(1,2,3)$ and $0^\circ < A,B,C < 180^\circ$, there is only one permutation of angles $(A,B,C)$ where the tangent of each angle results in a permutation of $(1,2,3)$. Therefore, all of the triangles in set $S$ are similar.

See Also

2006 Indonesia MO (Problems)
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed by
Problem 4
All Indonesia MO Problems and Solutions
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