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.
, then either travels [list] units vertically (up or down) and 
units horizontally (left or right); or units horizontally (left or right) and units vertically (up or down)., the ant can choose to go to one of eight possible points.
and 
, the ant can travel from 
to 
using at most 
moves.
be a set of 
elements and 
are subsets of (
), such that every one of them has at least 
elements.
and 
, with 
, such that the number of common elements of 
and 
is greater or equal to: 
, the sides 
and 
are parallel and the angles 
and 
are acute. Show that it is possible to divide the triangle 
into 4 disjoint triangle 
and the triangle 
into 4 disjoint triangles 
such that the triangles 
and 
are congruent for all .
has side lengths , , and 
. Select a point 
inside , and construct the incenters of 
, 
, and 
and denote them as 
, 
, 
. What is the maximum area of the triangle 
?
be non-negative real numbers such that 
Prove that
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of are either all odd or all even. is interesting if 2018 divides 
(the number of positive divisors of ). Determine all positive integers such that there exists an infinite arithmetic progression with common difference whose terms are all interesting.
with its center 
. There are two fixed points 
and one moving point 
on . The midpoint of the line segment 
is 
. 
is a fixed point on . Line 
intersects
at 
, and line 
intersects 
at 
. such that is always tangent to 
when varies.
such that:![\[
a^2 + b^2 \mid a^3 + b^3
\]](http://latex.artofproblemsolving.com/a/1/d/a1ddbd20c854e291da5670de449229d1664fa56c.png)
be an isosceles triangle with 
. The circle 
, passing through 
and 
, intersects segment at 
. The circle 
is tangent to at and passes through 
. Let and 
be the midpoints of segments and 
, respectively. The line 
intersects and at points and 
, respectively, where and are the intersections closer to . Prove that 
.
, where O is the circumcenter and I is the incenter of triangle ABC. In other words, 
. Since 
, this becomes 
. Thus we have
.
. It worked, but definitely not as nicely.
. It worked, but definitely not as nicely.
. It worked, but definitely not as nicely.
That's also what I did in the contest... Though it is not nice at all, it took me not much time to solve this problem.
, 
and 
. Let the incenter of the triangle be 
and let the point at which the incircle is tangent to be 
. Now note that![\[a=2R\cdot \sin{A} = 2R \cdot \sin{\frac{A}{2}} \cdot \cos{\frac{A}{2}}\]](http://latex.artofproblemsolving.com/d/4/a/d4ace78944ea3ec255b06277a4651857f5dd47bd.png)
,![\[AD=r\cdot \frac{\cos{\frac{A}{2}}}{\sin{\frac{A}{2}}}.\]](http://latex.artofproblemsolving.com/4/0/8/408741e1b5feef9662c66de2d07834f57db55d47.png)
yields that![\[\frac{b+c-a}{a}=\frac{2AD}{a}=\frac{r}{R\cdot \sin^2{\frac{A}{2}}}.\]](http://latex.artofproblemsolving.com/5/a/e/5ae2b0a6e75022f8d355af5a3c7a9a3d13448535.png)
, it follows that![\[\frac{MN^2}{a^2}=\frac{(b-a)^2 + (c-a)^2 -2(c-a)(b-a) \cos{A}}{a^2}\]](http://latex.artofproblemsolving.com/9/b/6/9b6a23683ec87c98f0ec9e76557815d6a8617162.png)
![\[\frac{MN^2}{a^2}=\frac{(b^2 + c^2 - 2bc\cos{A})-a(1-\cos{A})(b+c-a)}{a^2}\]](http://latex.artofproblemsolving.com/0/c/d/0cd57cbdbde37636c004aa22bd89378964a7b57d.png)
![\[\frac{MN^2}{a^2}=1-\frac{(1-\cos{A})(b+c-a)}{a}\]](http://latex.artofproblemsolving.com/a/e/8/ae88a71de241504906fe6f2da4cbde769ee875e5.png)
![\[\frac{MN^2}{a^2}=1-\frac{2\sin^2{\frac{A}{2}}(b+c-a)}{a}\]](http://latex.artofproblemsolving.com/0/4/0/0406eb1b23b797ccd19e33d80e2127fff88386cf.png)
![\[\frac{MN^2}{a^2}=1-\frac{2r}{R}.\]](http://latex.artofproblemsolving.com/b/4/0/b4079b520784c3247f5ad7061072485e0f902237.png)
![\[\frac{MN}{BC}=\sqrt{1-\frac{2r}{R}}.\]](http://latex.artofproblemsolving.com/3/1/6/316d2ed9d6e27548867f188108db869b41e35359.png)
of 
be . with rotation angle 
carries 
and 
.
, 
becomes equal to 
. Therefore the ratio becomes equal to 
. in terms of 
and . Then apply cosine rule on triangles 
and to get the value of in terms of . Now just expand both the sides and they become equal.
. By the problem conditions, 
and 
. Then, 
, and thus
From here, it is clear that
Because ![$$[ABC]=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}=\frac{rs}{2},$$](http://latex.artofproblemsolving.com/f/0/e/f0e4f05d668a0fa0e18f467ed307c9146a17e749.png)
it follows that 
Similarly, because ![$$[ABC]=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}=\frac{abc}{4R},$$](http://latex.artofproblemsolving.com/5/5/a/55a14a94973ae78d1426a294b7be98b409bcfe8e.png)
we can rewrite as 
As a result, 
and 
.
such that
.
in terms of 
,
,
.
,
,
,
![$\frac{1}{abc} \cdot [3abc-(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)+a^3+b^3+c^3]$](http://latex.artofproblemsolving.com/0/7/2/07209efe2e89337fda28d19fedcb93190d7081d0.png)
.
and 
for ease of notation, we have that:
.
and 
,
,
Notice 
and 
so 
Hence, 
and 
Notice ![$$\sqrt{s(s-a)(s-b)(s-c)}=[ABC]=\frac{abc}{4R}=sr$$](http://latex.artofproblemsolving.com/3/4/c/34c77bbea6fd7a3eace9eb63ebe4fb3ed8184b20.png)
so 
Therefore, 