[/quote]
,
be an integer. Prove that there exists a set 
of 
positive integers satisfying the following property: For every 
the set 
.
,
,
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
,
is an integer. Prove that there is a positive integer 
such that 
for all 
.
be a positive integer. On the table, we have 
ornaments in 
different colours.
grid of real numbers. Determine the greatest possible number of cells 
in the grid such that the entry in 
,
has side lengths 
,
,
inside 
,
,
and denote them as 
,
,
.
?
.
be a point on side 
,
be a point on the extension of 
such that 
.
be the circumcenter of 
,
satisfying 
.
be a prime and suppose 
is prime. Prove that 
is prim
the orthocenter of an acute triangle 
.
,
,
,
respectively such that ![\[ OD + DH = OE + EH = OF + FH\]](http://latex.artofproblemsolving.com/a/8/7/a874af4cb802b9b341b055140a08e096aaf2112e.png)
and the lines 
,
,
are concurrent.
Y by AdithyaBhaskar, Davi-8191, pragna0527, jhu08, TheCollatzConjecture, mathmax12, mathematicsy, HWenslawski, megarnie, son7, OlyMathSpirit, MathLuis, ImSh95, Adventure10, Mango247, lian_the_noob12, ItsBesi, Tastymooncake2, ehuseyinyigit, Rounak_iitr, cubres, and 5 other users
Let 
be a triangle with circumcentre 
. The points 
and 
are interior points of the sides 
and 
respectively. Let 
and 
be the midpoints of the segments 
and 
. respectively, and let 
be the circle passing through and . Suppose that the line is tangent to the circle . Prove that 
Proposed by Sergei Berlov, Russia
be a triangle with circumcentre 
. The points 
and 
are interior points of the sides 
and 
respectively. Let 
and 
be the midpoints of the segments 
and 
. respectively, and let 
be the circle passing through and . Suppose that the line is tangent to the circle . Prove that 
Proposed by Sergei Berlov, Russia
This post has been edited 2 times. Last edited by orl, Jul 15, 2009, 11:04 PM
AND 
and since 
,
.
.
is parallel to 
.
are middle lines in triangles 
and 
,
.
and 
are similar. Using side ratios, we get![\[ \frac{AP}{AQ}=\frac{MK}{ML}=\frac{BQ}{PC},\]](http://latex.artofproblemsolving.com/c/1/6/c1651c3ed4cdc0fc14f866cca7d0247019f907ff.png)
.
and apply Stewart's theorem for cevian 
![\[ OQ^2 = R^2 \cdot \frac{BQ}{AB} + R^2 \cdot \frac{AQ}{AB} - AQ \cdot BQ = R^2 - AQ \cdot BQ.\]](http://latex.artofproblemsolving.com/f/9/3/f93f84517cce9b7cd838fe0e0e2f5439d1210a2c.png)
![\[ OP^2 = R^2 - AP \cdot PC,\]](http://latex.artofproblemsolving.com/7/c/2/7c2fee9aa62b77c7f7ab5e9ea1d64b4c9afb268d.png)
.
.
.
.
if you take the circumcircle of ABC for the unit circle... But it's really a lot of computation.
.
,
,
maybe it would have given some space for interpretation (since triangles are more likely to be considered degenerated in some cases).
points?
and 
from the similarity of 
(i don't remember the order of the points) we get that 
(or something like this, i don't remember well), or equivalently 
.
and 
we get that we have to prove that
,
is the second intersection of 
,
is the perpendicular bisector of the line segment 
of 
.
be the reflection of 
across 
and 
the reflection of 
across 
.
to show that 
.
,
.
,
.
be the circumcircle of 
is tangent to 
so, 
and 
.
So, 
and 
(By Similarity).
Since 
and 
(Proved in the lemma), we get
.
is parallel to 
and that 
is parallel to 
.
and 
by similar triangles. We will use this later. Note also that 
and 
from the two sets of parallel lines. Thus, by AA similarity 
.
. Therefore, 
,
have the same power with respect to 
.
. 
because 
Then, since 
and 
are all obviously on 
are the midpoints of 
and 
respectively, it follows that 
so it then follows that
Then, since 
we have 
so 
Analogously, we have 
so it follows by AA Similarity that 
as desired. 
Therefore, we have
However, in this last equation, note that both the left-hand side and the right-hand side are equal to 
because of the facts that 
and 
respectively, so we are done. 
,
are midpoint of 
respectively, thus by converse of Thales theorem, 
and 
and
therefore 
and
Now, in 
is the midpoint of 
and 
.
and 
.
that is 
and 
thus
,
.
,
.
![\[\measuredangle MKL = \measuredangle PML = \measuredangle QPA, \quad \measuredangle MLK = \measuredangle QMK = \measuredangle PQA\]](http://latex.artofproblemsolving.com/4/6/d/46df5918c87b9a7a9718d19551090e84c07c9914.png)
represent the complex number 
.![\[\frac{\overline p - \overline a}{\overline q - \overline a} = \frac{k - m}{\ell - m} = \frac{\frac{b+p}{2} - \frac{p+q}{2}}{\frac{c+q}{2} - \frac{p+q}{2}} = \frac{b-q}{c-p}\]](http://latex.artofproblemsolving.com/e/b/9/eb9a25888f4e7eb020e8368b25b4100ddb057051.png)
![\[\implies (c \overline p - \overline ac + \overline ap) - |p|^2 = (b \overline q - \overline ab + \overline aq) - |q|^2.\]](http://latex.artofproblemsolving.com/c/5/3/c53d485366b8e7eed1e79a5085c0eb41397d595b.png)
of the unit circle gives us![\[a+c = p + ac \overline p \implies 1 + \overline ac = \overline ap + c \overline p,\]](http://latex.artofproblemsolving.com/3/c/d/3cdfc8873465e476055de1c0e681471c99d2f150.png)
![\[1 - |p|^2 = 1 - |q|^2 \implies OP = OQ \quad \blacksquare\]](http://latex.artofproblemsolving.com/0/1/b/01b3b6ed6b50edd8167595c2b9f478200f18cf16.png)
application.
Kb limit
and 
.
and 
,
.
.
and 
,
or equivalently 
.
)
From this it we immediately get 
and 
is similar to 
, since 
and 
.
,
. Now since P, Q have the same power, they must be equidistant from the center 
is oppositely similar to 
.
by the Tangent Lemma. Since 
,
.
,
Similarly, 
By AA similarity, the lemma is proven.
due to the similar triangles. Multiplying both sides by 
(since 
and 
,
is the circumradius of 
.
.
.
.
have sam power wrt 
![[asy]
import graph;
size(300);
defaultpen(fontsize(10));
// Define triangle vertices
pair A = (0,0), B = (8,0), C = (2,6);
draw(A--B--C--cycle, linewidth(0.8));
// Circumcenter O of triangle ABC (for an acute triangle, computed manually)
pair O = (4,2);
draw(circumcircle(A,B,C), dashed+blue);
// Parameters chosen to satisfy the tangency condition
real t = 0.8255; // for point P on CA
real s = 0.1; // for point Q on AB
// Define points P and Q
pair P = (2*t, 6*t); // P lies on CA
pair Q = (8*s, 0); // Q lies on AB
// Define midpoints: K (of BP), L (of CQ), and M (of PQ)
pair K = (B + P)/2;
pair L = (C + Q)/2;
pair M = (P + Q)/2;
// Compute circumcenter of circle Γ through K, L, M
// Note: K and M share the same y-coordinate, so their perpendicular bisector is vertical.
real bisectorX = (K.x + M.x)/2;
pair midKM = (bisectorX, K.y);
// Perpendicular bisector of LM:
pair midLM = ((L.x + M.x)/2, (L.y + M.y)/2);
real slopeLM = (L.y - M.y)/(L.x - M.x); // should equal 3
real perpSlope = -1/slopeLM; // slope of perpendicular bisector of LM
// The intersection of the vertical line x = bisectorX with the bisector of LM gives the center of Γ.
real centerGammaY = midLM.y + perpSlope*(bisectorX - midLM.x);
pair centerGamma = (bisectorX, centerGammaY);
real rGamma = abs(M - centerGamma);
// Draw the constructed elements
draw(P--Q, red+1.2); // line PQ (tangent to Γ at M)
draw(circle(centerGamma, rGamma), green+1.2); // circle Γ
draw(B--P, dotted);
draw(C--Q, dotted);
// Mark all key points
dot("$A$", A, SW, red);
dot("$B$", B, SE, yellow);
dot("$C$", C, N, green);
dot("$O$", O, S, heavygreen);
dot("$P$", P, NE, magenta);
dot("$Q$", Q, S, cyan);
dot("$K$", K, NE, orange);
dot("$L$", L, NW, green);
dot("$M$", M, S, blue);
[/asy]](http://latex.artofproblemsolving.com/b/d/9/bd9ffeb99058b455801fa2c1d1f4a89332195a10.png)
are equal. Note that that implies the problem condition since that means that
First of all, observe that 
is the 
m
,
- and, equivalently, we have 
.
which follows from
and
Now, the similiarity of the two triangles yields
As already stated, however, 
,
which directly implies 
,
Therefore 
Similarly 
so 
Now, note that 
By the triangle similarity we have 
Also by the Midpoint Theorem 
Thus these two fractions are equal and the desired result holds. QED
.
.![\[\angle KLM = \angle QMK = \angle AQM = \angle AQP.\]](http://latex.artofproblemsolving.com/8/e/0/8e0ee31d5f26c9b847a73a1116554d508cb150fe.png)
.![\[ \frac{QB}{PC} = \frac{MK}{ML} = \frac{AP}{AQ},\]](http://latex.artofproblemsolving.com/5/8/7/587d70db50d61080e8e240188e0e9629a26b2e34.png)
so 
,
