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: | ||
+ | |||
+ | 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. | ||
+ | |||
+ | ~Di Xu | ||
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Latest revision as of 17:32, 22 November 2023
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
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.
~Di Xu