AoPS Academy
and 
be two points inside triangle 
such that![\[ \angle MAB = \angle NAC\quad \mbox{and}\quad \angle MBA = \angle NBC.
\]](http://latex.artofproblemsolving.com/2/0/4/20410178c70693881f348c67454f01057a3fc40c.png)
![\[ \frac {AM \cdot AN}{AB \cdot AC} + \frac {BM \cdot BN}{BA \cdot BC} + \frac {CM \cdot CN}{CA \cdot CB} = 1.
\]](http://latex.artofproblemsolving.com/b/b/6/bb64f80eccb4634b9c9e15cade2f7c0b87b60923.png)
the set of all positive real numbers.
which have the property:![\[f(x)f(y)=2f(x+yf(x))\]](http://latex.artofproblemsolving.com/4/f/f/4ffbde35ff082585a188c1a4368d39338d1b07a4.png)
and 
.
+1 w
questions, each with 
answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?
, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
has only postive integers in forms
where 
![$f:[1,b]\rightarrow \mathbb{R},\ f(x)=x+\dfrac{bc}{x}$](http://latex.artofproblemsolving.com/5/6/d/56d1aa809cfad6667896f3ec4e2a1951b13a394c.png)
. As 
, we infer that ![$f(x)\ge f(b)=b+c,\ \forall x\in [1,b]$](http://latex.artofproblemsolving.com/3/9/6/396ded9b7014efff81a94e281d94b95dc71b62ab.png)
; in particular, 
.
, hence 
. It is easy to see that for 
, if 
, then 
. If 
, 
, so 
, which is in contradiction with the fact that 
. Thereby, 
.
. If 
, contradiction; so 
and 
, or equivalently 
and 
is odd, we get that 
. Furthermore, 
, hence 
. Now 
which together with the fact that 
, we get that 
, so the family of the solutions 
is described by the set
, then using 
, and 
, we get 
, which is clearly impossible. Thus, 
must hold.
and implies 
. Keeping this in mind, we now observe that, 
, implies 
. Thus, 
. Now, note also that, if 
, then 
, and thus, the largest power of 
dividing either 
or 
is exactly . Clearly, 
, and thus, 
. Moreover, if 
, then we again have a contradiction, as 
. Thus, 
. This yields, 
, which brings us, 
. Now, using 
, we immediately obtain 
. This immediately establishes (since is odd), that .
It follows that 
and 
we get 
Now 
and is an odd multiple of 2, we have that 4 must not divide either one of them. It follows that 
or 
But 
So 
Q.E.D.
with odd positive integers 
such that 
and 
. It is easy to see that 
. Further,
Of course, numbers 
are even. If both 
and 
are divisible by , then 
are not divisible by , so 
, that's impossible. Similar things happen when both 
and 
are divisible by , so we have two cases:
. So 
, hence 
. Further, 
, therefore 
, and we finally get 
.
. Here 
, hence . Also we have that 
, therefore and again.
and thus . Let
where 
are odd integers. Then, the condition becomes![\[2^{2k-2} - x^2 = 2^{2m-2} - y^2 \Longrightarrow 2^{2k-2} - 2^{2m-2} = x^2-y^2\]](http://latex.artofproblemsolving.com/4/d/d/4ddafb7c4a2ed27635fd75217fca16fcc439da28.png)
.
Note that![\[k\geq v_2(x^2-y^2) = v_2(2^{2k-2} - 2^{2m-2}) = 2m-2\]](http://latex.artofproblemsolving.com/2/2/a/22aa55b0d3cbecc2c70081b403a9f8361021af84.png)
where the first inequality is true because 
odd implies that
, and 
implies that 
. 
with equality when 
.
Note that![\[2^{2k-2} - 2^{2m-2} = x^2-y^2 \leq (2^{k-1}-1)^2 - 1^2,\]](http://latex.artofproblemsolving.com/f/8/b/f8b20fd1b76a3e6c7b65073e836c05410fbc908d.png)
thus 
, so . , and so the 
equality case must hold and we're done. 
.
:, we have: 
Therefore 
or . Note that: 
Note that 
are odd integers.
have a difference of 
which is an odd multiple of , we must have either: or 
. Notice that: 
which implies . Notice that: 
which implies . Also, we have: 
. Thus: 
Since is odd integer, and we are done.
. So, 
. Note that 
, and 
.
, 
, a contradiction as 
. So, 
and 
. Since 
, 
and 
. Hence, 
, and 
.
, consider a prime 
such that 
. 
, a contradiction, so 
.
be odd integers such that 
and 
and 
for some integers 
and 
,
or 
, contradiction )