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.
be a convex quadrilateral inscribed in a circle and satisfying 
. Points 
and 
are chosen on sides 
and 
such that 
and 
. Prove that 
.
be an integer. Ivan writes the numbers 
each on different cards. He then shuffles these 
cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.
integers 
, each less than or equal to 
, prove that at least one of them is divisible by some other member of the set.
is tangent to the circle 
at 
. and 
cut the circle at and respectively. 
is the in-centre of 
, and is on the line 
. 
and 
intersect at 
, while 
and 
intersect at 
. Prove that the points 
lie on a circle.
be the number of positive divisors of 
. Let 
be the number of positive divisors of which have remainders 
when divided by 
. Find all positive integral values of the fraction 
.
kids in a class. Is it true that for all positive integers 
it is possible to create several clubs each with 3 kids such that any pair of kids are both present in exactly one club?
has a right angle at 
. The centre of the circumcircle is called 
, and the base point of the normal from to is called . The point lies on 
with 
. The angle bisector of 
meets 
in 
. The lines 
and 
intersect in . Show that the lines 
and 
are parallel.
let 
denote the number of (positive) divisors of 
. Find all functions 
with the following properties: [list][*] 
for all 
.
divides 
for all 
, 
.[/list] be a triangle with circumcenter and orthocenter 
. 
the altitudes of . A point 
lies on line 
such that 
. A point 
lies on the circumcircle of such that 
are concurrent. 
meets 
at 
. Prove that 
are concyclic.
be a triangle. Distinct points , , lie on sides , , and , respectively, such that quadrilaterals 
and 
are cyclic. Line 
meets the circumcircle of again at . Let denote the reflection of across . Show that lies on the circumcircle of 
.
with side length 
is rotated about its center by angle 
, where 
, to form 
. The area of hexagon 
is 
. What is 
?![[asy]
defaultpen(fontsize(13)); size(200);
pair O=(0,0),A=dir(225),B=dir(-15),C=dir(105),D=rotate(38.21,O)*A,E=rotate(38.21,O)*B,F=rotate(38.21,O)*C;
draw(A--B--C--A,gray+0.4);draw(D--E--F--D,gray+0.4); draw(A--D--B--E--C--F--A,black+0.9); dot(O); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F));
[/asy]](http://latex.artofproblemsolving.com/8/6/7/86730ada3e1c6c3b65b2dda0ae37fb4eecc00ee7.png)
![[asy]
size(250);
import geometry;
// Define the side length of the equilateral triangle
real sideLength = 14;
real radius = sideLength / (2 * sin(60 * pi / 180)); // Radius of the circumscribed circle
// Define the rotation angle theta where tan(theta) = 5sqrt(3)/11
real theta = atan(5 * sqrt(3) / 11);
// Define the center of the circle
pair O = (0,0);
// Draw the circle in black
draw(circle(O, radius), linewidth(0.8) + black);
// Define the vertices of the original triangle, with A in the top left
pair A = dir(150) * radius; // Position A in the top-left
pair C = dir(270) * radius; // C is at the bottom
pair B = dir(30) * radius; // B is at the top-right
// Draw the original triangle in black
draw(A--B--C--cycle, linewidth(0.8) + black);
// Label the vertices of the original triangle in black
label("$A$", A, NW, black);
label("$B$", B, NE, black);
label("$C$", C, S, black);
// Rotate the triangle by angle -theta for a clockwise rotation
pair A1 = rotate(-degrees(theta)) * A;
pair B1 = rotate(-degrees(theta)) * B;
pair C1 = rotate(-degrees(theta)) * C;
// Draw the rotated triangle in black
draw(A1--B1--C1--cycle, linewidth(0.8) + black);
// Label the vertices of the rotated triangle in black
label("$A'$", A1, NW, black);
label("$B'$", B1, NE, black);
label("$C'$", C1, S, black);
// Connect the specified pairs of vertices
draw(A--C1, dashed + black);
draw(C--B1, dashed + black);
draw(B--A1, dashed + black);
// Connect each corresponding vertex between the original and rotated triangle
draw(A--A1, dashed + black);
draw(B--B1, dashed + black);
draw(C--C1, dashed + black);
// Add the center point O
dot(O, black);
label("$O$", O, NE, black);
// Project points A and A' down onto the horizontal line
pair A_proj = (A.x, 0);
pair A1_proj = (A1.x, 0);
// Draw only the segment OA_proj on the horizontal line
draw(O--A_proj, dashed + black);
// Project A and A' onto the horizontal line with dotted lines
draw(A--A_proj, dotted + black);
draw(A1--A1_proj, dotted + black);
// Mark the projections
dot(A_proj, black);
dot(A1_proj, black);
// Label the projections
label("$A_{proj}$", A_proj, S, black);
label("$A'_{proj}$", A1_proj, S, black);
// Add lines from O to A and O to A'
draw(O--A, dotted + black);
draw(O--A1, dotted + black);
// Find the intersection of line A'A'_{proj} and AB
pair X = intersectionpoint(A1--A1_proj, A--B);
// Draw and label the intersection point X
dot(X, black);
label("$X$", X, S, black);
[/asy]](http://latex.artofproblemsolving.com/b/6/5/b659aeec1480b38d0026756ffbb0c7f079d36d80.png)
. Calculation of the tangents of those angles and then an application of the tangent subtraction formula yields 
, 
, then the area of 
is 
. By Sine area formula, 
, and by LoC (angle = 120 by cyclic quad properties), 
. Then, solving these equations gives 
, so 
. Then, by trig bash, you get 
.
be the center of triangles and 
, be the foot of the altitude from to 
, and 
be the midpoint of . Remark that the condition 
implies that lies on segment 
(as opposed to segment 
). Furthermore, observe that is the center of circle 
, so by the Inscribed Angle Theorem 
. It suffices to compute 
and then use the Tangent Double Angle Formula to obtain .![[asy]
import olympiad;
size(200);
defaultpen(linewidth(0.6)+fontsize(11));
real t = 48;
pair A = dir(90), B = dir(210), C = dir(330), D = dir(90 + t), E = dir(210 + t), F = dir(330 + t), O = origin, P = foot(D,A,B), M = (A+B)/2;
draw(D--P--A^^rightanglemark(D,P,A,2));
draw(A--B--C--cycle^^D--E--F--cycle^^A--O--D--B);
draw(unitcircle,linetype("3 3"));
label("$A$",A,dir(O--A));
label("$B$",B,dir(O--B));
label("$C$",C,dir(O--C));
label("$D$",D,dir(O--D));
label("$E$",E,dir(O--E));
label("$F$",F,dir(O--F));
label("$O$",O,S);
label("$P$",P,SE);
[/asy]](http://latex.artofproblemsolving.com/3/5/7/357ec3fcbf08dd5272592b803bf8b39a54711034.png)
is 
. Because hexagon consists of plus three copies of 
, the area of is 
. This means 
. Additionally, because 
, 
and 
. It follows from the Pythagorean Theorem that
Finally, 
, so 
and![\[
\tan \theta = \tan(2\angle DOB) = \frac{\frac{2\sqrt 3}5}{1 - \frac{3}{25}} = \frac{5\sqrt 3}{11}.
\]](http://latex.artofproblemsolving.com/2/c/6/2c643648627deccfdc121ccd4715c66da922eb5b.png)
be the center of triangles and , be the foot of the altitude from to , and be the midpoint of . Remark that the condition implies that lies on segment (as opposed to segment ). Furthermore, observe that is the center of circle , so by the Inscribed Angle Theorem . It suffices to compute and then use the Tangent Double Angle Formula to obtain . is . Because hexagon consists of plus three copies of , the area of is . This means . Additionally, because , and . It follows from the Pythagorean Theorem thatFinally, , so and / area to use for this problem, given that the solution is pretty agnostic to the choice of ?
/ area to use for this problem, given that the solution is pretty agnostic to the choice of ?
by writing 
. I wanted 
to be an integer, because otherwise 
would be some annoying mixed radical and the computation would spiral during the double angle step at the end.
terms, because those naturally appear when working with equilateral triangles (no funny business here!).
. (That is, I wanted the configuration to be "general" enough so that people actually needed to work to find the answer. Otherwise, it would feel as if the problem was set up for no good reason.)
. Then 
and 
. Furthermore, 
so the altitude from in is 
. Then ![$[ADB]=\frac{196}{\sqrt 3}\sin(\gamma)\sin(60-\gamma)$](http://latex.artofproblemsolving.com/8/2/5/825b39e72720fadf50c35bfca74ba77bb3c21c14.png)
. Since ![$3[ADB]+[ABC]=91\sqrt 3$](http://latex.artofproblemsolving.com/0/e/8/0e8820cdd1335d93564d99967d26ff90b2febeb5.png)
, ![$[ADB]=\frac{42}{\sqrt 3}$](http://latex.artofproblemsolving.com/6/c/b/6cb0f98a1540d4b61d6d1f64360b3eb469cf68d6.png)
, so 
. Then 
, so ![\[\cos(60-\theta) = \frac{13}{14}.\]](http://latex.artofproblemsolving.com/7/e/c/7ec14507b90b3a8a577007d1713cee887a107014.png)
This also gives 
, so 
. Finally, ![\[\tan(\theta) = \tan(60-(60-\theta)) = \frac{\sqrt 3 - \frac{3\sqrt 3}{13}}{1 + \frac 9{13}}=\frac{10\sqrt 3}{22} = \boxed{\frac{5\sqrt 3}{11}}.\]](http://latex.artofproblemsolving.com/8/b/5/8b5379aed100475fdbc8f12e52615f91e522fba7.png)
.
.![$[AOD] + [DOB]$](http://latex.artofproblemsolving.com/d/8/3/d83bd6e9e0aeb31a8d02a9a757ba2710bdedb378.png)
.
,
.
.
.
.
.
so 
.
.