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

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==Solution==
 
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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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If the phrase "maximum value" in the original problem is changed to "least upper bound of", then the problem should have the solution above.  
 
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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----Di Xu
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~Di Xu
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*[[Mock AIME 1 2006-2007 Problems/Problem 13 | Previous Problem]]
 
*[[Mock AIME 1 2006-2007 Problems/Problem 13 | Previous Problem]]

Latest revision as of 16:32, 22 November 2023

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

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.

~Di Xu

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