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Difference between revisions of "2012 AMC 12B Problems/Problem 25"
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== Solution 2 ==
 
== Solution 2 ==
 
This is just another way for the reasoning of solution 1. Define a "cell" to be a rectangle in the set of <math>S.</math> For example, a cell can be  
 
This is just another way for the reasoning of solution 1. Define a "cell" to be a rectangle in the set of <math>S.</math> For example, a cell can be  
−
size(200,200,IgnoreAspect);
+
<math>size(200,200,IgnoreAspect);
 
real f(real t) {return t;}
 
real f(real t) {return t;}
 
draw(graph(f,0,10),red);
 
draw(graph(f,0,10),red);
 
pen thin=linewidth(0.5*linewidth());
 
pen thin=linewidth(0.5*linewidth());
−
xaxis("<math>x</math>",BottomTop,grey,LeftTicks(begin=false,end=false,extend=true,
+
xaxis("</math>x<math>",BottomTop,grey,LeftTicks(begin=false,end=false,extend=true,
 
                                 ptick=thin));
 
                                 ptick=thin));
−
yaxis("<math>y</math>",LeftRight,grey,RightTicks(begin=false,end=false,extend=true,
+
yaxis("</math>y<math>",LeftRight,grey,RightTicks(begin=false,end=false,extend=true,
−
                                 ptick=thin));
+
                                 ptick=thin));</math>
  
 
==Video Solution by Richard Rusczyk==
 
==Video Solution by Richard Rusczyk==

Revision as of 17:38, 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. Define a "cell" to be a rectangle in the set of $S.$ For example, a cell can be $size(200,200,IgnoreAspect); real f(real t) {return t;} draw(graph(f,0,10),red); pen thin=linewidth(0.5*linewidth()); xaxis("$x$",BottomTop,grey,LeftTicks(begin=false,end=false,extend=true, ptick=thin)); yaxis("$y$",LeftRight,grey,RightTicks(begin=false,end=false,extend=true, ptick=thin));$

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12b/279

~dolphin7

See Also

2012 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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