AoPS Academy
![\[ x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. \]](http://latex.artofproblemsolving.com/5/6/b/56b8cb8c13616501dd8308989cea2b338e91598b.png)
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)
(i dont remember exactly if it was 8032038 or some other number, so correct me on this)
(i posted this at 12:29) lol
?
.
.![$[AOD] + [DOB]$](http://latex.artofproblemsolving.com/d/8/3/d83bd6e9e0aeb31a8d02a9a757ba2710bdedb378.png)
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so 
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