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Difference between revisions of "2024 AIME II Problems/Problem 15"
Line 35: Line 35:
 
dot(A12);
 
dot(A12);
 
</asy>
 
</asy>
 +
 +
==Solution 1==
 +
By Furaken
 +
 +
There are two kinds of such rectangles: those whose sides are parallel to some edges of the regular 12-gon (Case 1, and those whose sides are not (Case 2).
 +
 +
For Case 1, WLOG assume that the rectangle's sides are horizontal and vertical (don't forget to multiply by 3 at the end of Case 1). Then the rectangle's sides coincide with these segments as shown in the diagram.
 +
<asy>
 +
real r = pi/6;
 +
pair A1 = (cos(r),sin(r));
 +
pair A2 = (cos(2r),sin(2r));
 +
pair A3 = (cos(3r),sin(3r));
 +
pair A4 = (cos(4r),sin(4r));
 +
pair A5 = (cos(5r),sin(5r));
 +
pair A6 = (cos(6r),sin(6r));
 +
pair A7 = (cos(7r),sin(7r));
 +
pair A8 = (cos(8r),sin(8r));
 +
pair A9 = (cos(9r),sin(9r));
 +
pair A10 = (cos(10r),sin(10r));
 +
pair A11 = (cos(11r),sin(11r));
 +
pair A12 = (cos(12r),sin(12r));
 +
dot(A1);
 +
dot(A2);
 +
dot(A3);
 +
dot(A4);
 +
dot(A5);
 +
dot(A6);
 +
dot(A7);
 +
dot(A8);
 +
dot(A9);
 +
dot(A10);
 +
dot(A11);
 +
dot(A12);
 +
pair B1 = (0.5,0.5);
 +
pair B2 = (-0.5,0.5);
 +
pair B3 = (-0.5,-0.5);
 +
pair B4 = (0.5,-0.5);
 +
dot(B1);
 +
dot(B2);
 +
dot(B3);
 +
dot(B4);
 +
draw(A1--A5--A7--A11--cycle);
 +
draw(A2--A4--A8--A10--cycle);
 +
draw(A3--A9);
 +
draw(A6--A12);
 +
label("$A_1$", A1, NE);
 +
label("$A_2$", A2, NE);
 +
label("$A_3$", A3, N);
 +
label("$A_4$", A4, NW);
 +
label("$A_5$", A5, NW);
 +
label("$A_6$", A6, W);
 +
label("$A_7$", A7, SW);
 +
label("$A_8$", A8, SW);
 +
label("$A_9$", A9, S);
 +
label("$A_{10}$", A10, SE);
 +
label("$A_{11}$", A11, SE);
 +
label("$A_{12}$", A12, E);
 +
label("$B_1$", B1, SW);
 +
label("$B_2$", B2, SE);
 +
label("$B_3$", B3, NE);
 +
label("$B_4$", B4, NW);
 +
</asy>
 +
We use inclusion-exclusion for this. There are 30 valid rectangles contained in <math>A_1A_5A_7A_{11}</math>, as well as 30 in <math>A_2A_4A_8A_{10}</math>. However, the 9 rectangles contained in <math>B_1B_2B_3B_4</math> have been counted twice, so we subtract 9 and we have 51 rectangles in the diagram. Multiplying by 3, we get 153 rectangles for Case 1.
 +
 +
For Case 2, we have this diagram.
 +
<asy>
 +
real r = pi/6;
 +
pair A1 = (cos(r),sin(r));
 +
pair A2 = (cos(2r),sin(2r));
 +
pair A3 = (cos(3r),sin(3r));
 +
pair A4 = (cos(4r),sin(4r));
 +
pair A5 = (cos(5r),sin(5r));
 +
pair A6 = (cos(6r),sin(6r));
 +
pair A7 = (cos(7r),sin(7r));
 +
pair A8 = (cos(8r),sin(8r));
 +
pair A9 = (cos(9r),sin(9r));
 +
pair A10 = (cos(10r),sin(10r));
 +
pair A11 = (cos(11r),sin(11r));
 +
pair A12 = (cos(12r),sin(12r));
 +
dot(A1);
 +
dot(A2);
 +
dot(A3);
 +
dot(A4);
 +
dot(A5);
 +
dot(A6);
 +
dot(A7);
 +
dot(A8);
 +
dot(A9);
 +
dot(A10);
 +
dot(A11);
 +
dot(A12);
 +
draw(A1--A6--A7--A12--cycle);
 +
draw(A3--A4--A9--A10--cycle);
 +
draw(A2--A5--A8--A11--cycle);
 +
label("$A_1$", A1, NE);
 +
label("$A_2$", A2, NE);
 +
label("$A_3$", A3, N);
 +
label("$A_4$", A4, NW);
 +
label("$A_5$", A5, NW);
 +
label("$A_6$", A6, W);
 +
label("$A_7$", A7, SW);
 +
label("$A_8$", A8, SW);
 +
label("$A_9$", A9, S);
 +
label("$A_{10}$", A10, SE);
 +
label("$A_{11}$", A11, SE);
 +
label("$A_{12}$", A12, E);
 +
</asy>
 +
There are 36 rectangles contained within <math>A_2A_5A_8A_{11}</math>, and 18 that use points outside <math>A_2A_5A_8A_{11}</math>. So we get a total of <math>3(36+18)=162</math> rectangles for Case 2.
 +
 +
Adding the two cases together, we get the answer <math>\boxed{315}</math>.
  
 
==See also==
 
==See also==

Revision as of 18:33, 8 February 2024

Problem

Find the number of rectangles that can be formed inside a fixed regular dodecagon where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles.

[asy] unitsize(60); real r = pi/6; pair A1 = (cos(r),sin(r)); pair A2 = (cos(2r),sin(2r)); pair A3 = (cos(3r),sin(3r)); pair A4 = (cos(4r),sin(4r)); pair A5 = (cos(5r),sin(5r)); pair A6 = (cos(6r),sin(6r)); pair A7 = (cos(7r),sin(7r)); pair A8 = (cos(8r),sin(8r)); pair A9 = (cos(9r),sin(9r)); pair A10 = (cos(10r),sin(10r)); pair A11 = (cos(11r),sin(11r)); pair A12 = (cos(12r),sin(12r)); draw(A1--A2--A3--A4--A5--A6--A7--A8--A9--A10--A11--A12--cycle); filldraw(A3--A2--A9--A8--cycle, mediumgray, linewidth(1.2)); draw(A5--A12); dot(0.365*A4); dot(0.365*A1); dot(A1); dot(A2); dot(A3); dot(A4); dot(A5); dot(A6); dot(A7); dot(A8); dot(A9); dot(A10); dot(A11); dot(A12); [/asy]

Solution 1

By Furaken

There are two kinds of such rectangles: those whose sides are parallel to some edges of the regular 12-gon (Case 1, and those whose sides are not (Case 2).

For Case 1, WLOG assume that the rectangle's sides are horizontal and vertical (don't forget to multiply by 3 at the end of Case 1). Then the rectangle's sides coincide with these segments as shown in the diagram. [asy] real r = pi/6; pair A1 = (cos(r),sin(r)); pair A2 = (cos(2r),sin(2r)); pair A3 = (cos(3r),sin(3r)); pair A4 = (cos(4r),sin(4r)); pair A5 = (cos(5r),sin(5r)); pair A6 = (cos(6r),sin(6r)); pair A7 = (cos(7r),sin(7r)); pair A8 = (cos(8r),sin(8r)); pair A9 = (cos(9r),sin(9r)); pair A10 = (cos(10r),sin(10r)); pair A11 = (cos(11r),sin(11r)); pair A12 = (cos(12r),sin(12r)); dot(A1); dot(A2); dot(A3); dot(A4); dot(A5); dot(A6); dot(A7); dot(A8); dot(A9); dot(A10); dot(A11); dot(A12); pair B1 = (0.5,0.5); pair B2 = (-0.5,0.5); pair B3 = (-0.5,-0.5); pair B4 = (0.5,-0.5); dot(B1); dot(B2); dot(B3); dot(B4); draw(A1--A5--A7--A11--cycle); draw(A2--A4--A8--A10--cycle); draw(A3--A9); draw(A6--A12); label("$A_1$", A1, NE); label("$A_2$", A2, NE); label("$A_3$", A3, N); label("$A_4$", A4, NW); label("$A_5$", A5, NW); label("$A_6$", A6, W); label("$A_7$", A7, SW); label("$A_8$", A8, SW); label("$A_9$", A9, S); label("$A_{10}$", A10, SE); label("$A_{11}$", A11, SE); label("$A_{12}$", A12, E); label("$B_1$", B1, SW); label("$B_2$", B2, SE); label("$B_3$", B3, NE); label("$B_4$", B4, NW); [/asy] We use inclusion-exclusion for this. There are 30 valid rectangles contained in $A_1A_5A_7A_{11}$, as well as 30 in $A_2A_4A_8A_{10}$. However, the 9 rectangles contained in $B_1B_2B_3B_4$ have been counted twice, so we subtract 9 and we have 51 rectangles in the diagram. Multiplying by 3, we get 153 rectangles for Case 1.

For Case 2, we have this diagram. [asy] real r = pi/6; pair A1 = (cos(r),sin(r)); pair A2 = (cos(2r),sin(2r)); pair A3 = (cos(3r),sin(3r)); pair A4 = (cos(4r),sin(4r)); pair A5 = (cos(5r),sin(5r)); pair A6 = (cos(6r),sin(6r)); pair A7 = (cos(7r),sin(7r)); pair A8 = (cos(8r),sin(8r)); pair A9 = (cos(9r),sin(9r)); pair A10 = (cos(10r),sin(10r)); pair A11 = (cos(11r),sin(11r)); pair A12 = (cos(12r),sin(12r)); dot(A1); dot(A2); dot(A3); dot(A4); dot(A5); dot(A6); dot(A7); dot(A8); dot(A9); dot(A10); dot(A11); dot(A12); draw(A1--A6--A7--A12--cycle); draw(A3--A4--A9--A10--cycle); draw(A2--A5--A8--A11--cycle); label("$A_1$", A1, NE); label("$A_2$", A2, NE); label("$A_3$", A3, N); label("$A_4$", A4, NW); label("$A_5$", A5, NW); label("$A_6$", A6, W); label("$A_7$", A7, SW); label("$A_8$", A8, SW); label("$A_9$", A9, S); label("$A_{10}$", A10, SE); label("$A_{11}$", A11, SE); label("$A_{12}$", A12, E); [/asy] There are 36 rectangles contained within $A_2A_5A_8A_{11}$, and 18 that use points outside $A_2A_5A_8A_{11}$. So we get a total of $3(36+18)=162$ rectangles for Case 2.

Adding the two cases together, we get the answer $\boxed{315}$.

See also

2024 AIME II (Problems • Answer Key • Resources)
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