AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.![\[ \begin{tabular}{c c c c}
{} & \textbf{Adams} & \textbf{Baker} & \textbf{Adams and Baker} \\
\textbf{Boys:} & 71 & 81 & 79 \\
\textbf{Girls:} & 76 & 90 & ? \\
\textbf{Boys and Girls:} & 74 & 84 & \\
\end{tabular}
\]](http://latex.artofproblemsolving.com/8/5/c/85c75770c3f90af49df0db600dead2a29e6674b3.png)
, such that: 
For all 
.
be positive real numbers such that : 
. Prove that : 
be an odd integer. Prove that there exists a prime number 
such that![\[
p \mid 2^{\varphi(n)} - 1 \quad \text{but} \quad p \nmid n.
\]](http://latex.artofproblemsolving.com/3/5/1/3514b64ecd5f7b2586d5bfeae59b7c9504d5b455.png)
be a triangle for which there exists an interior point 
such that 
. Let the lines 
and 
meet the sides 
and 
at 
and 
respectively. Prove that ![\[ AB+AC\geq4DE. \]](http://latex.artofproblemsolving.com/3/a/4/3a49c5f8bd2930adca7bde2601dd045d4bf05ff6.png)
such that ![\[f(xf(y)+f(x+y)+1)=(y+1)f(x+1)\]](http://latex.artofproblemsolving.com/4/9/e/49ec422b640766de653d9d640bbc45065a6ac375.png)
holds for all 
.
on one variable with real coefficients such that the equation 
has exactly 
ordered pairs 
as solutions?
such that
for all nonnegative real numbers 
, 
, and 
with 
.
such that any real numbers 
and 
satisfy
Proposed by EmilXM
. be the unique point such that 
is an isosceles trapezoid with 
. Then, triangle 
is isosceles, so we must have 
, and the conclusion follows from Ptolemy's Theorem.
, 
, and 
. The answer is 
, or C.
be the side lengths. Note that the cosines of all the angles are rational (LOC), so taking cosines of both sides, we have 
, so ![\[ \cos(\angle B) + 1 = \frac{a^2 + c^2 - b^2}{2ac} + 1= \frac{(a+c)^2 - b^2) }{2ac}= \frac{(a+b+c)(a + c - b)}{2ac} \]](http://latex.artofproblemsolving.com/0/8/2/0821df23edf4e1bd06b2bc973da347f8b08213ca.png)
must be two times a square. Multiplying this by 
gives that ![\[ f(a,b,c) = \frac{(a+b+c)(a + c - b)}{ac} \]](http://latex.artofproblemsolving.com/2/d/a/2da4bd22273efa2aa4c75a64b446903d27b75adc.png)
is the square of a rational number.
is an odd prime dividing 
so that 
is odd, then divides 
.
, which is a contradiction since 
is the square of a rational. 
doesn't work because it requires one of to be a multiple of and 
doesn't work because it requires one of to be a multiple of 
(thus triangle inequality won't hold), so 
. Now, we see that 
must divide one of and 
must divide one of , so we could have 
or 
.
of gives that can't divide 
, so 
and 
is 
, meaning 
. This is impossible as the sum of the angles would exceed 
.
.
isn't the smallest in the triangle, we cannot have 
. If 
, then 
, which is not a square. Therefore, 
and 
, which is a square. Then, 
, so 
(must be positive as 
), which can be verified to be true as 
.
works, so it is possible for the perimeter to be 
.
-bisector hit at . Then 
and 
. From the angle condition, 
is isosceles so 
. Now use Stewart with 
as the cevian:![\[ \frac{ab^3c}{(a+c)^2} + \frac{b^3c^2}{(a+c)^2} = \frac{a^2bc + c^2ba}{a+c}\]](http://latex.artofproblemsolving.com/5/7/2/57239b5ac5a0348d1fd47423a6ea0df5f8a2e72f.png)
![\[\iff ab^3c + b^3c^2 = (a+c)^2abc\]](http://latex.artofproblemsolving.com/9/a/d/9ad742deaf55840a7dfb7cd5dbe4f865a9a11324.png)
![\[\iff (a+c)b^2 = (a+c)^2a\]](http://latex.artofproblemsolving.com/a/c/c/acce7f6cc20cad3a685e99a0296de1d3f9315f45.png)
![\[\iff b^2 = a(a+c)\]](http://latex.artofproblemsolving.com/1/a/f/1afc7d879a17972f532ac5950a3c2ee7affc5547.png)
now just check stuff to find that 4,5,6 works -> C. screwed this up by being messing up random division and stuff when simplifying the stewart equation and thus getting ridiculous diophantines. skull
, 
, and 
, which when expanded give![\[
\sin(\theta) : 2\sin(\theta)\cos(\theta) : 3\sin(\theta) - 4\sin^3(\theta)
\]](http://latex.artofproblemsolving.com/7/4/0/7400a3db9a71c52aac7a0b253688fa401d7ee4a9.png)
or![\[
1 : 2\cos(\theta) : 3 - 4\sin^2(\theta)
\]](http://latex.artofproblemsolving.com/7/e/7/7e743be1c0b598d90163858c4a6771b24b017b30.png)
must be rational; otherwise, it's not possible to have integer side lengths for the triangle (and rational makes 
rational as well). Now, we just test some values of .
fails because it makes the last term 0, 
fails, 
gives 
, which gives 
, or 
, too large.
fails because it makes the last term negative, but 
gives 
triangle.
and above, the fraction denominator becomes large, and it's impossible to have a smaller perimeter than 
.
, , and , which when expanded giveor must be rational; otherwise, it's not possible to have integer side lengths for the triangle (and rational makes rational as well). Now, we just test some values of . fails because it makes the last term 0, fails, gives , which gives , or , too large. fails because it makes the last term negative, but gives triangle. and above, the fraction denominator becomes large, and it's impossible to have a smaller perimeter than .
.
for coprime and 
, the primitive triangle would be 
, for a perimeter of 
. We must have 
, so the minimal solution is given by 
, 
for 15. This is the 4-5-6 triangle.
for coprime 
and 
, the primitive triangle would be 
, for a perimeter of 
. Now 
So the minimal solution is given by and , for a perimeter of 28.
triangle. "I'm here to tell you all about 4-5-6 triangle facts. Did you know I'm the minimal scalene triangle which can be used to form a disphenoid? Did you know my area is 
? Did you know that I'm the minimal triangle with one angle twice the other angle? Did you know my perimeter is 
?" It told me this and two more questions, but I was barely conscious. I asked the queer floating triangle, "What do you want from me? Why are you telling me this?" The triangle told me "Thrice on the MAA tests will you see a triangle. If you mess up all three times, I will smite you and rend you to dust." triangle that I won't be able to solve. Help! What do I do?
to meet 
at . Let 
, 
, 
as usual, and let 
, so 
. Now 
as 
, and 
, so 
by AA.
gives us 
.
, so 
. Rearranging gives 
so 
.: clearly needs to be composite to avoid 
or 
. Trying gives us 
, which gives us a degenerate solution of 
. Now when , we get 
, which works and gives us 
.
, which would require 
, so we have achieved our minimum.
such that 
be a triangle with integer side lengths and the property that 
.
meet the circle at 
,
.
