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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==
+
== 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 S</math> that have one vertex on the line <math>y=0</math> do not reflect to a triangle <math>\subseteq S</math>. So we only need to look at right triangles that have vertices <math>(0,5), (*,0), (*,*)</math>. There are three cases:
+
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:
  
 
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=(2,4), 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 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 23:28, 2 August 2024

Problem 25

Let $S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}$. Let $T$ be the set of all right triangles whose vertices are in $S$. For every right triangle $t=\triangle{ABC}$ with vertices $A$, $B$, and $C$ in counter-clockwise order and right angle at $A$, let $f(t)=\tan(\angle{CBA})$. What is \[\prod_{t\in T} f(t)?\]

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24}$

Solution 1

Consider reflections. For any right triangle $ABC$ with the right labeling described in the problem, any reflection $A'B'C'$ labeled that way will give us $\tan CBA \cdot \tan C'B'A' = 1$. First we consider the reflection about the line $y=2.5$. Only those triangles $\subseteq T$ that have one vertex at $(0,5)$ do not reflect to a traingle $\subseteq T$. Within those triangles, consider a reflection about the line $y=5-x$. Then only those triangles $\subseteq T$ that have one vertex on the line $y=0$ do not reflect to a triangle $\subseteq T$. So we only need to look at right triangles that have vertices $(0,5), (*,0), (*,*)$. There are three cases:

Case 1: $A=(0,5)$. Then $B=(*,0)$ is impossible.

Case 2: $B=(0,5)$. Then we look for $A=(x,y)$ such that $\angle BAC=90^{\circ}$ and that $C=(*,0)$. They are: $(A=(x,5), C=(x,0))$, $(A=(3,2), C=(1,0))$ and $(A=(4,1), C=(3,0))$. The product of their values of $\tan \angle CBA$ is $\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}$.

Case 3: $C=(0,5)$. Then $A=(*,0)$ is impossible.

Therefore $\boxed{\textbf{(B)} \ \frac{625}{144}}$ 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 $S.$ 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 $x$ and the shorter side to be $y,$ then the product will be $\frac{x}{y} \cdot \frac{y}{x} \cdot \frac{x}{y} \cdot \frac{y}{x}=1,$ 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 $(0,5), (*,0), (*,*),$ where $*$ is a number defined by the boundaries of $S.$ Thus, by the three cases, our answer is $\boxed{\textbf{(B)} \ \frac{625}{144}}.$

~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

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