Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 14"
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==Solution== | ==Solution== | ||
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+ | |||
+ | 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>\\ | ||
+ | Proof of lemma: \\ | ||
+ | Construct <math>PH\perp{AB}</math> at <math>H</math>. \\ | ||
+ | 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>\\\\ | ||
+ | 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?\\ | ||
+ | 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. | ||
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Revision as of 13:57, 4 December 2021
Problem
Three points ,
, and
are fixed such that
lies on segment
, closer to point
. Let
and
where
and
are positive integers. Construct circle
with a variable radius that is tangent to
at
. Let
be the point such that circle
is the incircle of
. Construct
as the midpoint of
. Let
denote the maximum value
for fixed
and
where
. If
is an integer, find the sum of all possible values of
.
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: \\
Proof of lemma: \\
Construct
at
. \\
Case (i)
\\
\\\\
Case (ii)
\\
\\\\
Case (iii)
\\
, proof done.\\\\
Now we try to find . \\
Let O be the centre of the incircle, and
be the inradius.\\
\\\\
\\\\
Similarly,
\\\\
Therefore,
\\\\
Therefore,
.\\\\
Therefore, all possible values of
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,
is when "=" is achieved. However, in this case,
, 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.