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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>
  
−
[[2012 AMC 12B Problems/Problem 25|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 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 1==
+
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=(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.
 +
 
 +
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);
  
−
Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that  <math>\tan x  \tan(90^{\circ}-x) = \tan x  \cot x = 1.</math>
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draw(A--B--C--cycle);
 +
draw(anglemark(C,B,A, 3999));
 +
</asy>
  
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Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.
+
<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>
  
−
So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangle <math>tan B = \frac{y}{x}</math>.
+
<asy>
−
Multiplying them all together gives: <math>\frac{1}{1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot  \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot  \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{4} \cdot  \frac{4}{1} \cdot \frac{4}{2} \cdot \frac{4}{3} \cdot \frac{4}{4} \cdot  \frac{5}{1} \cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} = \boxed{\textbf{(E)} \ \frac{625}{24}. } </math>
+
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>
  
−
==Solution 2==
+
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.
−
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 S</math> that have one vertex at <math>(0,5)</math> do not reflect to a traingle <math>\subseteq S</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:
 
  
−
Case 1: <math>A=(0,5)</math>. Then <math>B=(*,0)</math> is impossible.
+
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
  
−
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>.
+
~dolphin7
  
−
Case 3: <math>C=(0,5)</math>. Then <math>A=(*,0)</math> is impossible.
+
== See Also ==
  
−
Therefore <math>\boxed{\textbf{(B)} \frac{625}{144}}</math> is the answer.
+
{{AMC12 box|year=2012|ab=B|num-b=24|after=Last Problem}}
 +
{{MAA Notice}}

Revision as of 20:03, 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); draw(A--B--C--cycle); draw(anglemark(C,B,A, 3999)); [/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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