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Difference between revisions of "1966 IMO Problems/Problem 2"

(→‎Solution)
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Let <math>a</math>, <math>b</math>, and <math>c</math> be the lengths of the sides of a triangle, and <math> \alpha,\beta,\gamma </math> respectively, the angles opposite these sides. If,
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Let <math>a</math>, <math>b</math>, and <math>c</math> be the lengths of the sides of a triangle, and <math> \alpha,\beta,\gamma </math> respectively, the angles opposite these sides. Prove that if
  
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<cmath> a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}) </cmath>
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<cmath> a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}), </cmath>
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the triangle is isosceles.
  
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Prove that the triangle is isosceles.
 
  
 
==Solution==
 
==Solution==
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We'll prove that the triangle is isosceles with <math>a=b</math>.
 
We'll prove that the triangle is isosceles with <math>a=b</math>.
 
We'll prove that <math>a=b</math>. Assume by way of contradiction WLOG that <math>a>b</math>.
 
We'll prove that <math>a=b</math>. Assume by way of contradiction WLOG that <math>a>b</math>.
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Now, <math>\pi >\alpha -\beta >0</math> (as <math>\alpha+\beta +\gamma = \pi</math> and the angles are positive), <math>\tan \frac{\alpha -\beta}{2}\neq 0</math>, and furthermore, <math>\tan \frac{\alpha+\beta}{2}>0</math>. By all the above, <cmath>\left(a-b+\tan \frac{\alpha +\beta}{2}(a\tan\alpha -b\tan \beta) \right)>0</cmath>
 
Now, <math>\pi >\alpha -\beta >0</math> (as <math>\alpha+\beta +\gamma = \pi</math> and the angles are positive), <math>\tan \frac{\alpha -\beta}{2}\neq 0</math>, and furthermore, <math>\tan \frac{\alpha+\beta}{2}>0</math>. By all the above, <cmath>\left(a-b+\tan \frac{\alpha +\beta}{2}(a\tan\alpha -b\tan \beta) \right)>0</cmath>
 
Which contradicts our assumption, thus <math>a\leq b</math>. By the symmetry of the condition, using the same arguments, <math>a\geq b</math>. Hence <math>a=b</math>.
 
Which contradicts our assumption, thus <math>a\leq b</math>. By the symmetry of the condition, using the same arguments, <math>a\geq b</math>. Hence <math>a=b</math>.
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==Solution 2==
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First, we'll prove that both <math>\alpha</math> and <math>\beta</math> are acute.
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At least one of them has to be acute because these are angles
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of a triangle.  We can assume that <math>\alpha</math> is acute.  We want
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to show that <math>\beta</math> is acute as well.  For a proof by
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contradiction, assume <math>\beta \ge \frac{\pi}{2}</math>.
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From the hypothesis, it follows that
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<math>(a + b) \tan \frac{\alpha + \beta}{2} = a \tan \alpha + b \tan \beta</math>.
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From <math>\alpha < \beta</math> it follows that <math>a < b</math>.  So,
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(Solution by pf02, September 2024)
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TO BE CONTINUED.  SAVING MID WAY SO I DON'T LOOSE WORK DONE SO FAR.
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==See Also==
 
==See Also==
 
{{IMO box|year=1966|num-b=1|num-a=3}}
 
{{IMO box|year=1966|num-b=1|num-a=3}}

Revision as of 00:18, 29 September 2024

Let $a$, $b$, and $c$ be the lengths of the sides of a triangle, and $\alpha,\beta,\gamma$ respectively, the angles opposite these sides. Prove that if

\[a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}),\]

the triangle is isosceles.


Solution

We'll prove that the triangle is isosceles with $a=b$. We'll prove that $a=b$. Assume by way of contradiction WLOG that $a>b$. First notice that as $\gamma = \pi -\alpha-\beta$ then and the identity $\tan\left(\frac \pi 2 - x \right)=\cot x$ our equation becomes: \[a+b=\cot \frac{\alpha +\beta}{2}\left(a\tan \alpha + b\tan \beta \right)\]\[\iff a\tan\frac{\alpha +\beta}{2}+b\tan \frac{\alpha +\beta}{2}=a\tan \alpha + b\tan \beta\] \[\iff a\left(\tan \alpha -\tan \frac{\alpha +\beta}{2}\right)+b\left(\tan \beta -\tan \frac{\alpha +\beta}{2} \right)=0\] Using the identity $\tan (A-B)=\frac {\tan A-\tan B}{1+\tan A\tan B}$ $\iff \tan A-\tan B=\tan(A-B)(1+\tan A\tan B)$ and inserting this into the above equation we get: \[\iff a\tan \frac{\alpha -\beta}{2}\left(1+\tan \alpha \tan \frac{\alpha +\beta}{2}\right)+b\tan \frac{\beta -\alpha}{2}\left(1+\tan \beta \tan \frac{\alpha +\beta}{2} \right)=0\] \[\underbrace{\iff}_{\tan -A=-\tan A}a\tan \frac{\alpha -\beta}{2}\left(1+\tan \alpha \tan \frac{\alpha +\beta}{2}\right)-b\tan \frac{\alpha -\beta}{2}\left(1+\tan \beta \tan \frac{\alpha +\beta}{2} \right)=0\] \[\iff \tan \frac{\alpha -\beta}{2}\left(a-b+\tan \frac{\alpha +\beta}{2}(a\tan\alpha -b\tan \beta) \right)=0\] Now, since $a>b$ and the definitions of $a,b,\alpha,\beta$ being part of the definition of a triangle, $\alpha >\beta$. Now, $\pi >\alpha -\beta >0$ (as $\alpha+\beta +\gamma = \pi$ and the angles are positive), $\tan \frac{\alpha -\beta}{2}\neq 0$, and furthermore, $\tan \frac{\alpha+\beta}{2}>0$. By all the above, \[\left(a-b+\tan \frac{\alpha +\beta}{2}(a\tan\alpha -b\tan \beta) \right)>0\] Which contradicts our assumption, thus $a\leq b$. By the symmetry of the condition, using the same arguments, $a\geq b$. Hence $a=b$.


Solution 2

First, we'll prove that both $\alpha$ and $\beta$ are acute. At least one of them has to be acute because these are angles of a triangle. We can assume that $\alpha$ is acute. We want to show that $\beta$ is acute as well. For a proof by contradiction, assume $\beta \ge \frac{\pi}{2}$.

From the hypothesis, it follows that $(a + b) \tan \frac{\alpha + \beta}{2} = a \tan \alpha + b \tan \beta$.

From $\alpha < \beta$ it follows that $a < b$. So,




(Solution by pf02, September 2024)

TO BE CONTINUED. SAVING MID WAY SO I DON'T LOOSE WORK DONE SO FAR.


See Also

1966 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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