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Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 14"

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Please see below an attempted solution to understand why this problem doesn't have a solution:
 
Please see below an attempted solution to understand why this problem doesn't have a solution:
  
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Lemma: <math>\cot{B}-\cot{A}=2\cot{\angle{AMP}}</math>\\
+
Lemma: <math>\cot{B}-\cot{A}=2\cot{\angle{AMP}}</math>
−
Proof of lemma: \\
+
 
−
Construct <math>PH\perp{AB}</math> at <math>H</math>.  \\
+
Proof of lemma:  
−
Case (i) <math>A<90^\circ</math>\\
+
 
−
<math>\cot{B}-\cot{A}=\dfrac{BH}{PH}-\dfrac{AH}{PH}=\dfrac{BM+MH}{PH}-\dfrac{AM-MH}{PH}=\dfrac{BM+MH-AM+MH}{PH}=\dfrac{2MH}{PH}=\cot{\angle{AMP}}</math>\\\\
+
Construct <math>PH\perp{AB}</math> at <math>H</math>.   
−
Case (ii) <math>A>90^\circ</math>\\
+
 
−
<math>\cot{B}-\cot{A}=\dfrac{BH}{PH}+\dfrac{AH}{PH}=\dfrac{BM+MH}{PH}+\dfrac{MH-AM}{PH}=\dfrac{BM+MH-AM+MH}{PH}=\dfrac{2MH}{PH}=\cot{\angle{AMP}}</math>\\\\
+
Case (i) <math>A<90^\circ</math>
−
Case (iii) <math>A=90^\circ</math>\\
+
 
−
<math>\cot{B}-\cot{A}=\dfrac{BA}{PA}-0=\dfrac{2MA}{PA}=\cot{\angle{AMP}}</math>,  proof done.\\\\
+
<math>\cot{B}-\cot{A}=\dfrac{BH}{PH}-\dfrac{AH}{PH}=\dfrac{BM+MH}{PH}-\dfrac{AM-MH}{PH}=\dfrac{BM+MH-AM+MH}{PH}=\dfrac{2MH}{PH}=\cot{\angle{AMP}}</math>
 +
 
 +
Case (ii) <math>A>90^\circ</math>
 +
 
 +
<math>\cot{B}-\cot{A}=\dfrac{BH}{PH}+\dfrac{AH}{PH}=\dfrac{BM+MH}{PH}+\dfrac{MH-AM}{PH}=\dfrac{BM+MH-AM+MH}{PH}=\dfrac{2MH}{PH}=\cot{\angle{AMP}}</math>
 +
 
 +
Case (iii) <math>A=90^\circ</math>
 +
 
 +
<math>\cot{B}-\cot{A}=\dfrac{BA}{PA}-0=\dfrac{2MA}{PA}=\cot{\angle{AMP}}</math>,  proof done.
 +
 
 +
 
 +
Now we try to find <math>f(m,n)</math>.
 +
 
 +
Let O be the centre of the incircle,  and <math>r</math> be the inradius.
 +
 
 +
<math>\tan{\angle{OAB}} = \dfrac{OT}{AT} = \dfrac{r}{m}</math>
 +
 
 +
<math>\tan{\angle{PAB}} = \tan{(2\angle{OAB})} =  \dfrac{2\tan{\angle{OAB}}}{1-\tan^2{\angle{OAB}}} = \dfrac{2r/m}{1-(r/m)^2} = \dfrac{2mr}{m^2-r^2}</math>
 +
 
 +
Similarly,  <math>\tan{\angle{PBA}} = \dfrac{2nr}{n^2-r^2}</math>
 +
 
 +
Therefore,  <math>\tan^2{\angle{AMP}} = \dfrac{1}{\cot^2{\angle{AMP}}} = \dfrac{1}{\dfrac{(\cot{\angle{PBA}}-\cot{\angle{PAB}})^2}{2^2}} = \dfrac{4}{(\cot{\angle{PBA}}-\cot{\angle{PAB}})^2} = \dfrac{4}{\Big(\dfrac{1}{\tan{\angle{PBA}}}-\dfrac{1}{\tan{\angle{PAB}}}\Big)^2} = \dfrac{4}{\Big(\dfrac{n^2-r^2}{2nr}-\dfrac{m^2-r^2}{2mr}\Big)^2} = \dfrac{16m^2n^2r^2}{[(n^2-r^2)m-(m^2-r^2)n]^2} = \dfrac{16m^2n^2r^2}{[(n-m)(r^2+mn)]^2} \le \dfrac{16m^2n^2r^2}{[(n-m)(2r\sqrt{mn})]^2} = \dfrac{4mn}{n-m}</math>
 +
 
 +
Therefore, <math>f(m,49)=\dfrac{196m}{49-m}</math>.
 +
 
 +
Therefore, all possible values of <math>m</math> are 48, 47, 42, 35, and the answer is 48+47+42+35=172.
 +
 
 +
What's the problem with this solution?
  
−
Now we try to find <math>f(m,n)</math>. \\
 
−
Let O be the centre of the incircle,  and <math>r</math> be the inradius.\\
 
−
<math>\tan{\angle{OAB}} = \dfrac{OT}{AT} = \dfrac{r}{m}</math>\\\\
 
−
<math>\tan{\angle{PAB}} = \tan{(2\angle{OAB})} =  \dfrac{2\tan{\angle{OAB}}}{1-\tan^2{\angle{OAB}}} = \dfrac{2r/m}{1-(r/m)^2} = \dfrac{2mr}{m^2-r^2}</math>\\\\
 
−
Similarly,  <math>\tan{\angle{PBA}} = \dfrac{2nr}{n^2-r^2}</math>\\\\
 
−
Therefore,  <math>\tan^2{\angle{AMP}} = \dfrac{1}{\cot^2{\angle{AMP}}} = \dfrac{1}{\dfrac{(\cot{\angle{PBA}}-\cot{\angle{PAB}})^2}{2^2}} = \dfrac{4}{(\cot{\angle{PBA}}-\cot{\angle{PAB}})^2} = \dfrac{4}{\Big(\dfrac{1}{\tan{\angle{PBA}}}-\dfrac{1}{\tan{\angle{PAB}}}\Big)^2} = \dfrac{4}{\Big(\dfrac{n^2-r^2}{2nr}-\dfrac{m^2-r^2}{2mr}\Big)^2} = \dfrac{16m^2n^2r^2}{[(n^2-r^2)m-(m^2-r^2)n]^2} = \dfrac{16m^2n^2r^2}{[(n-m)(r^2+mn)]^2} \le \dfrac{16m^2n^2r^2}{[(n-m)(2r\sqrt{mn})]^2} = \dfrac{4mn}{n-m}</math>\\\\
 
−
Therefore, <math>f(m,49)=\dfrac{196m}{49-m}</math>.\\\\
 
−
Therefore, all possible values of <math>m</math> are 48, 47, 42, 35, and the answer is 48+47+42+35=172.\\\\
 
−
What's the problem with this solution?\\
 
 
When AM-GM was used, <math>r=\sqrt{mn}</math> is when "=" is achieved. However, in this case, <math>PA\parallel{PB}</math>, so contradiction.  
 
When AM-GM was used, <math>r=\sqrt{mn}</math> is when "=" is achieved. However, in this case, <math>PA\parallel{PB}</math>, so contradiction.  
  

Revision as of 13:02, 4 December 2021

Problem

Three points $A$, $B$, and $T$ are fixed such that $T$ lies on segment $AB$, closer to point $A$. Let $AT=m$ and $BT=n$ where $m$ and $n$ are positive integers. Construct circle $O$ with a variable radius that is tangent to $AB$ at $T$. Let $P$ be the point such that circle $O$ is the incircle of $\triangle APB$. Construct $M$ as the midpoint of $AB$. Let $f(m,n)$ denote the maximum value $\tan^{2}\angle AMP$ for fixed $m$ and $n$ where $n>m$. If $f(m,49)$ is an integer, find the sum of all possible values of $m$.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Please see below an attempted solution to understand why this problem doesn't have a solution:

Lemma: $\cot{B}-\cot{A}=2\cot{\angle{AMP}}$

Proof of lemma:

Construct $PH\perp{AB}$ at $H$.

Case (i) $A<90^\circ$

$\cot{B}-\cot{A}=\dfrac{BH}{PH}-\dfrac{AH}{PH}=\dfrac{BM+MH}{PH}-\dfrac{AM-MH}{PH}=\dfrac{BM+MH-AM+MH}{PH}=\dfrac{2MH}{PH}=\cot{\angle{AMP}}$

Case (ii) $A>90^\circ$

$\cot{B}-\cot{A}=\dfrac{BH}{PH}+\dfrac{AH}{PH}=\dfrac{BM+MH}{PH}+\dfrac{MH-AM}{PH}=\dfrac{BM+MH-AM+MH}{PH}=\dfrac{2MH}{PH}=\cot{\angle{AMP}}$

Case (iii) $A=90^\circ$

$\cot{B}-\cot{A}=\dfrac{BA}{PA}-0=\dfrac{2MA}{PA}=\cot{\angle{AMP}}$, proof done.


Now we try to find $f(m,n)$.

Let O be the centre of the incircle, and $r$ be the inradius.

$\tan{\angle{OAB}} = \dfrac{OT}{AT} = \dfrac{r}{m}$

$\tan{\angle{PAB}} = \tan{(2\angle{OAB})} = \dfrac{2\tan{\angle{OAB}}}{1-\tan^2{\angle{OAB}}} = \dfrac{2r/m}{1-(r/m)^2} = \dfrac{2mr}{m^2-r^2}$

Similarly, $\tan{\angle{PBA}} = \dfrac{2nr}{n^2-r^2}$

Therefore, $\tan^2{\angle{AMP}} = \dfrac{1}{\cot^2{\angle{AMP}}} = \dfrac{1}{\dfrac{(\cot{\angle{PBA}}-\cot{\angle{PAB}})^2}{2^2}} = \dfrac{4}{(\cot{\angle{PBA}}-\cot{\angle{PAB}})^2} = \dfrac{4}{\Big(\dfrac{1}{\tan{\angle{PBA}}}-\dfrac{1}{\tan{\angle{PAB}}}\Big)^2} = \dfrac{4}{\Big(\dfrac{n^2-r^2}{2nr}-\dfrac{m^2-r^2}{2mr}\Big)^2} = \dfrac{16m^2n^2r^2}{[(n^2-r^2)m-(m^2-r^2)n]^2} = \dfrac{16m^2n^2r^2}{[(n-m)(r^2+mn)]^2} \le \dfrac{16m^2n^2r^2}{[(n-m)(2r\sqrt{mn})]^2} = \dfrac{4mn}{n-m}$

Therefore, $f(m,49)=\dfrac{196m}{49-m}$.

Therefore, all possible values of $m$ are 48, 47, 42, 35, and the answer is 48+47+42+35=172.

What's the problem with this solution?

When AM-GM was used, $r=\sqrt{mn}$ is when "=" is achieved. However, in this case, $PA\parallel{PB}$, so contradiction.

If the phrase "maximum value" in the original problem is changed to "least upper bound of", then the problem should have the solution above.

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