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Difference between revisions of "2024 AIME II Problems/Problem 15"

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(→‎Solution 1)
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Adding the two cases together, we get the answer <math>\boxed{315}</math>.
 
Adding the two cases together, we get the answer <math>\boxed{315}</math>.
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 +
==Solution 2==
 +
 +
Using the same diagram as Solution 1, we can get the number of rectangles from Case 1 by adding the number of rectangles of <math>A_2</math> <math>A_8</math> <math>A_8</math> <math>A_{10}</math> and <math>A_1</math> <math>A_5</math> <math>A_7</math> <math>A_{11}</math> and then subtracting the overlaps,
 +
 +
<cmath>\binom{5}{2}\binom{3}{2} + \binom{5}{2}\binom{3}{2} - \binom{3}{2}\binom{3}{2}</cmath>
 +
<cmath>=51</cmath>
 +
 +
We multiply this by 3 to get the total number of rectangles for Case 1, which is 153.
 +
 +
For Case 2, we can first get the total number of rectangles from
 +
<math>A_2A_3A_4A_5A_8A_9A_{10}A_{11}</math> then add <math>A_1A_6A_7A_{12}</math> and subtract by the overlaps,
 +
<cmath>\binom{4}{2}\binom{4}{2} + \binom{6}{2} - \binom{4}{2} + \binom{6}{2} - \binom{4}{2}</cmath>
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<cmath>= 54</cmath>
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Multiply that by 3 and add it to Case 1 to get <math>\boxed{315}</math>.
 +
 +
~pengf
  
 
==See also==
 
==See also==

Revision as of 12:13, 9 February 2024

Problem

Find the number of rectangles that can be formed inside a fixed regular dodecagon ($12$-gon) 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(0.6 inch); for(int i=0; i<360; i+=30) { dot(dir(i), 4+black); draw(dir(i)--dir(i+30)); } draw(dir(120)--dir(330)); filldraw(dir(210)--dir(240)--dir(30)--dir(60)--cycle, mediumgray, linewidth(1.5)); draw((0,0.366)--(0.366,0), linewidth(1.5)); [/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. To be honest, you can count the rectangles here in whatever way you like. [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 $\boxed{315}$.

Solution 2

Using the same diagram as Solution 1, we can get the number of rectangles from Case 1 by adding the number of rectangles of $A_2$ $A_8$ $A_8$ $A_{10}$ and $A_1$ $A_5$ $A_7$ $A_{11}$ and then subtracting the overlaps,

\[\binom{5}{2}\binom{3}{2} + \binom{5}{2}\binom{3}{2} - \binom{3}{2}\binom{3}{2}\] \[=51\]

We multiply this by 3 to get the total number of rectangles for Case 1, which is 153.

For Case 2, we can first get the total number of rectangles from $A_2A_3A_4A_5A_8A_9A_{10}A_{11}$ then add $A_1A_6A_7A_{12}$ and subtract by the overlaps, \[\binom{4}{2}\binom{4}{2} + \binom{6}{2} - \binom{4}{2} + \binom{6}{2} - \binom{4}{2}\] \[= 54\] Multiply that by 3 and add it to Case 1 to get $\boxed{315}$.

~pengf

See also

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