Difference between revisions of "2012 AMC 12B Problems/Problem 25"
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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} </math> | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} </math> | ||
− | + | == Solution 1 == | |
− | + | Consider reflections. For any right triangle <math>ABC</math> with the right labeling described in the problem, any reflection <math>A'B'C'</math> labeled that way will give us <math>\tan CBA \cdot \tan C'B'A' = 1</math>. First we consider the reflection about the line <math>y=2.5</math>. Only those triangles <math>\subseteq T</math> that have one vertex at <math>(0,5)</math> do not reflect to a traingle <math>\subseteq T</math>. Within those triangles, consider a reflection about the line <math>y=5-x</math>. Then only those triangles <math>\subseteq T</math> that have one vertex on the line <math>y=0</math> do not reflect to a triangle <math>\subseteq T</math>. So we only need to look at right triangles that have vertices <math>(0,5), (*,0), (*,*)</math>. There are three cases: | |
− | ==Solution== | ||
− | Consider reflections. For any right triangle <math>ABC</math> with the right labeling described in the problem, any reflection <math>A'B'C'</math> labeled that way will give us <math>\tan CBA \cdot \tan C'B'A' = 1</math>. First we consider the reflection about the line <math>y=2.5</math>. Only those triangles <math>\subseteq | ||
Case 1: <math>A=(0,5)</math>. Then <math>B=(*,0)</math> is impossible. | Case 1: <math>A=(0,5)</math>. Then <math>B=(*,0)</math> is impossible. | ||
− | Case 2: <math>B=(0,5)</math>. Then we look for <math>A=(x,y)</math> such that <math>\angle BAC=90^{\circ}</math> and that <math>C=(*,0)</math>. They are: <math>(A=(x,5), C=(x,0))</math>, <math>(A=(2 | + | Case 2: <math>B=(0,5)</math>. Then we look for <math>A=(x,y)</math> such that <math>\angle BAC=90^{\circ}</math> and that <math>C=(*,0)</math>. They are: <math>(A=(x,5), C=(x,0))</math>, <math>(A=(3,2), C=(1,0))</math> and <math>(A=(4,1), C=(3,0))</math>. The product of their values of <math>\tan \angle CBA</math> is <math>\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}</math>. |
Case 3: <math>C=(0,5)</math>. Then <math>A=(*,0)</math> is impossible. | Case 3: <math>C=(0,5)</math>. Then <math>A=(*,0)</math> is impossible. | ||
Therefore <math>\boxed{\textbf{(B)} \ \frac{625}{144}}</math> is the answer. | Therefore <math>\boxed{\textbf{(B)} \ \frac{625}{144}}</math> is the answer. | ||
+ | |||
+ | == Solution 2 == | ||
+ | This is just another way for the reasoning of solution 1. Picture the question to be a grid of unit squares instead of a coordinate system. Note that the restriction, (This is just to make visualization easier.) Define a "cell" to be a rectangle in the set of <math>S.</math> For example, a cell can be (labeled in a red): | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.5 cm); | ||
+ | draw((0,1)--(0,5),black); | ||
+ | draw((1,0)--(1,5),black); | ||
+ | draw((2,0)--(2,5),black); | ||
+ | draw((3,0)--(3,5),black); | ||
+ | draw((4,0)--(4,5),black); | ||
+ | draw((0,1)--(5,1),black); | ||
+ | draw((5,0)--(5,5),black); | ||
+ | draw((1,0)--(5,0),black); | ||
+ | draw((0,2)--(5,2),black); | ||
+ | draw((0,3)--(5,3),black); | ||
+ | draw((0,4)--(5,4),black); | ||
+ | draw((0,5)--(5,5),black); | ||
+ | draw((0,1)--(5,1),red); | ||
+ | draw((0,1)--(0,5),red); | ||
+ | draw((0,5)--(5,5),red); | ||
+ | draw((5,5)--(5,1),red); | ||
+ | </asy> | ||
+ | |||
+ | Note that choosing any three points and labeling them according to the problem will result in a product of one. For example, with the cell we just labeled, the four triangles we can create are: | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.5 cm); | ||
+ | draw((0,1)--(5,1),red); | ||
+ | draw((0,1)--(0,5),red); | ||
+ | draw((0,5)--(5,5),red); | ||
+ | draw((5,5)--(5,1),red); | ||
+ | draw((0,1)--(5,1),black); | ||
+ | draw((0,1)--(0,5),black); | ||
+ | draw((0,5)--(5,1),black); | ||
+ | |||
+ | pair A, B, C; | ||
+ | |||
+ | A = (0,1); | ||
+ | B = (0,5); | ||
+ | C = (5,1); | ||
+ | </asy> | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.5 cm); | ||
+ | draw((0,1)--(5,1),red); | ||
+ | draw((0,1)--(0,5),red); | ||
+ | draw((0,5)--(5,5),red); | ||
+ | draw((5,5)--(5,1),red); | ||
+ | draw((0,1)--(5,1),black); | ||
+ | draw((5,1)--(5,5),black); | ||
+ | draw((5,5)--(0,1),black); | ||
+ | </asy> | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.5 cm); | ||
+ | draw((0,1)--(5,1),red); | ||
+ | draw((0,1)--(0,5),red); | ||
+ | draw((0,5)--(5,5),red); | ||
+ | draw((5,5)--(5,1),red); | ||
+ | draw((0,5)--(5,5),black); | ||
+ | draw((5,1)--(5,5),black); | ||
+ | draw((0,5)--(5,1),black); | ||
+ | </asy> | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.5 cm); | ||
+ | draw((0,1)--(5,1),red); | ||
+ | draw((0,1)--(0,5),red); | ||
+ | draw((0,5)--(5,5),red); | ||
+ | draw((5,5)--(5,1),red); | ||
+ | draw((0,5)--(5,5),black); | ||
+ | draw((0,1)--(0,5),black); | ||
+ | draw((0,1)--(5,5),black); | ||
+ | </asy> | ||
+ | |||
+ | If we define the longer side to be <math>x</math> and the shorter side to be <math>y,</math> then the product will be <math>\frac{x}{y} \cdot \frac{y}{x} \cdot \frac{x}{y} \cdot \frac{y}{x}=1,</math> and we are done. | ||
+ | |||
+ | Otherwise, the three points are not contained in a "cell." This will result in the solution 1 path as described before. Our three points must take the form <math>(0,5), (*,0), (*,*),</math> where <math>*</math> is a number defined by the boundaries of <math>S.</math> Thus, by the three cases, our answer is <math>\boxed{\textbf{(B)} \ \frac{625}{144}}.</math> | ||
+ | |||
+ | ~wesserwes7254 | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2012amc12b/279 | ||
+ | |||
+ | ~dolphin7 | ||
== See Also == | == See Also == | ||
− | {{AMC12 box|year=2012|ab=B|num-b=24|after=}} | + | {{AMC12 box|year=2012|ab=B|num-b=24|after=Last Problem}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:28, 3 August 2024
Problem 25
Let .
Let
be the set of all right triangles whose vertices are in
. For every right triangle
with vertices
,
, and
in counter-clockwise order and right angle at
, let
. What is
Solution 1
Consider reflections. For any right triangle with the right labeling described in the problem, any reflection
labeled that way will give us
. First we consider the reflection about the line
. Only those triangles
that have one vertex at
do not reflect to a traingle
. Within those triangles, consider a reflection about the line
. Then only those triangles
that have one vertex on the line
do not reflect to a triangle
. So we only need to look at right triangles that have vertices
. There are three cases:
Case 1: . Then
is impossible.
Case 2: . Then we look for
such that
and that
. They are:
,
and
. The product of their values of
is
.
Case 3: . Then
is impossible.
Therefore is the answer.
Solution 2
This is just another way for the reasoning of solution 1. Picture the question to be a grid of unit squares instead of a coordinate system. Note that the restriction, (This is just to make visualization easier.) Define a "cell" to be a rectangle in the set of For example, a cell can be (labeled in a red):
Note that choosing any three points and labeling them according to the problem will result in a product of one. For example, with the cell we just labeled, the four triangles we can create are:
If we define the longer side to be and the shorter side to be
then the product will be
and we are done.
Otherwise, the three points are not contained in a "cell." This will result in the solution 1 path as described before. Our three points must take the form where
is a number defined by the boundaries of
Thus, by the three cases, our answer is
~wesserwes7254
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc12b/279
~dolphin7
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.