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 "Mock AIME 1 2006-2007 Problems/Problem 14"

(→‎Solution)
(→‎Solution)
 
(3 intermediate revisions by one other user not shown)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
−
{{solution}}
 
  
 
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:
  
−
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.  
  
 
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.  
 +
 +
~Di Xu
  
 
----
 
----

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_1_2006-2007_Problems/Problem_14&oldid=205259"