[/quote]
,
be reals such that 
and 
Prove that
with right angle 
,
is drawn. An arbitrary circle passing through points 
meets the segments 
,
and 
for the second time at points 
,
and 
respectively. Segments 
and 
meet at point 
.
or the sum of areas of triangles 
and 
?
and 
Prove that
is divisibly by 
.
and 
be two infinite sequences of integers. Prove that there exists an infinite sequence of integers 
such that for any positive integer 
,![\[
\sum_{k|n} k \cdot z_k^{\frac{n}{k}} = \left( \sum_{k|n} k \cdot x_k^{\frac{n}{k}} \right) \cdot \left( \sum_{k|n} k \cdot y_k^{\frac{n}{k}} \right).
\]](http://latex.artofproblemsolving.com/2/5/9/259ee32c89befe7a3ee2797e0d5b1df1c3066cf6.png)
such that for all reals 
Proposed by Aritra12, India
and 
imply that either 
or 
.
implies that 
.
also implies that 
implies that for any 
,
or 
.
,
is injective. 
and 
imply that 
for all 
.
.
such that 
,
gives 
for all such 
.
then gives 
and thus 
.
which means that 
.
for all 
for some 
,
which is impossible. Thus 
in this case.
let 
be the number whose 
-
-
.
.
and 
.
Prove 
Find maximum 
Prove that
Let 
Prove that
Prove that
and 
Prove that
Let 
Prove that
,
,
be the midpoints of sides BC, CA, and AB, respectively. Point 
is located on the ray 
so that 
is similar to 
.
and 
on rays 
and 
,
,
,
are concurrent, i.e. these three lines intersect at a point.
and a constant 
,
and 
are also friendly polynomials. Prove that 
.
Y by might_guy, Adventure10, Mango247, and 8 other users
Geometry is my weakest part although i have tried very hard to improve lately. I want to ask geometry experts of this forum : How do you train yourself on geometry ? What is your way of thinking when dealing with a hard geometry problem like 2008/6 or 2011/6 ? In my country, during national olimpiad and tst the geometry problems are usually easy or medium problems and i feel like we are not trained well enough. When i tried 2008/6 or 2011/6 i got stuck with no idea at all. Can you share what's your way of cracking it ? Did you deal with other problems with similar kind of diagram before so you can relate to it? (like in 2011/6 the fact that the the incenter of the determined triangle lies on the circumcircle of triangle ABC appears in previous test). Or did you try to draw a very good diagram and try to guess some properties ?
Also what books or problem sets do you think is appropriate for training ?
Sorry for this long post and my poor english
Also what books or problem sets do you think is appropriate for training ?
Sorry for this long post and my poor english
This post has been edited 2 times. Last edited by inge, Nov 27, 2020, 8:00 AM
(which amusingly came up as USA JMO 2014 #6) then the solution path is quite natural. If not, you are at a severe disadvantage. (Shameless plug: I'm in the process of trying to get a geometry book published (
,
.
and 
are. The diagram gives you an immediate, rapid way to disprove false conjectures, and also suggests new ones.
be the triangle formed by 
,
,
.
.
.
and 
look true as well. (I also discovered and proved the incenter lemma from my diagram, but I did not end up using it in my solution. However, it would certainly have been worth marks at the IMO.)![[asy]
size(8cm);
defaultpen(fontsize(8pt));
pair A = dir(110);
dot("$A$", A, dir(A));
pair B = dir(195);
dot("$B$", B, dir(160));
pair C = dir(325);
dot("$C$", C, 1.4*dir(30));
pair P = dir(270);
dot("$P$", P, dir(P));
draw(unitcircle);
draw(A--B--C--cycle, blue);
pair U = P+(2,0);
pair V = 2*P-U;
pair X_1 = reflect(B,C)*P;
pair Y_1 = reflect(C,A)*P;
pair Z_1 = reflect(A,B)*P;
pair X_2 = extension(B, C, U, V);
dot(X_2);
pair Y_2 = extension(C, A, U, V);
dot(Y_2);
pair Z_2 = extension(A, B, U, V);
dot(Z_2);
draw(B--Z_2, dotted+blue);
draw(C--Y_2, dotted+blue);
draw(C--X_2, dotted+blue);
draw(X_2--Z_2);
pair A_1 = extension(Y_1, Y_2, Z_1, Z_2);
dot("$A_1$", A_1, dir(A_1));
pair B_1 = extension(Z_1, Z_2, X_1, X_2);
dot("$B_1$", B_1, dir(B_1));
pair C_1 = extension(X_1, X_2, Y_1, Y_2);
dot("$C_1$", C_1, dir(50));
draw(A_1--B_1--C_1--cycle, green);
draw(C_1--X_2, dotted+green);
draw(circumcircle(A_1, B_1, C_1));
pair A_2 = A*A/P;
dot("$A_2$", A_2, dir(-20));
pair B_2 = B*B/P;
dot("$B_2$", B_2, dir(130));
pair C_2 = C*C/P;
dot("$C_2$", C_2, dir(C_2));
draw(A_2--B_2--C_2--cycle, red);
pair T = extension(A_1, A_2, B_1, B_2);
dot("$T$", T, dir(T));
draw(T--A_1, dashed);
draw(T--B_1, dashed);
draw(T--C_1, dashed);
[/asy]](http://latex.artofproblemsolving.com/8/5/c/85cee610e59798effe2c659c363c0f92f7870415.png)
defined are homothetic. So my conjecture is almost definitely true, and I simply have to show that 
,
,
concur on 
,
.
,
lying on the incircle.
