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Difference between revisions of "2015 AMC 8 Problems/Problem 19"

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== Problem ==
 
A triangle with vertices as <math>A=(1,3)</math>, <math>B=(5,1)</math>, and <math>C=(4,4)</math> is plotted on a <math>6\times5</math> grid. What fraction of the grid is covered by the triangle?
 
A triangle with vertices as <math>A=(1,3)</math>, <math>B=(5,1)</math>, and <math>C=(4,4)</math> is plotted on a <math>6\times5</math> grid. What fraction of the grid is covered by the triangle?
  
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<asy>
 
<asy>
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draw((1,0)--(1,5),linewidth(.5));
 
draw((1,0)--(1,5),linewidth(.5));
 
draw((2,0)--(2,5),linewidth(.5));
 
draw((2,0)--(2,5),linewidth(.5));
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</asy>
 
</asy>
  
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==Solution 1==
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==Solutions==
  
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The area of <math>\triangle ABC</math> is equal to half the product of its base and height.  By the Pythagorean Theorem, we find its height is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>, and its base is <math>\sqrt{2^2+4^2}=\sqrt{20}</math>.  We multiply these and divide by <math>2</math> to find the of the triangle is <math>\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5</math>.  Since the grid has an area of <math>30</math>, the fraction of the grid covered by the triangle is <math>\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}</math>.
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===Solution 1===
  
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==Solution 2==
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The area of <math>\triangle ABC</math> is equal to half the product of its base and height.  By the Pythagorean Theorem, we find its height is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>, and its base is <math>\sqrt{2^2+4^2}=\sqrt{20}</math>.  We multiply these and divide by <math>2</math> to find the area of the triangle is <math>\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5</math>.  Since the grid has an area of <math>30</math>, the fraction of the grid covered by the triangle is <math>\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}</math>.
−
Note angle <math>\angle ACB</math> is right, thus the area is <math>\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=10 \times \dfrac{1}{2}=5</math> thus the fraction of the total is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math>
 
  
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==Solution 3==
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===Solution 2===
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By the Shoelace theorem, the area of <math>\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12)|=|\dfrac{1}{2}(-10)|=5</math>.
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Note angle <math>\angle ACB</math> is right; thus, the area is <math>\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=10 \times \dfrac{1}{2}=5</math>; thus, the fraction of the total is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math>.
  
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This means the fraction of the total area is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math>
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===Solution 3===
 +
By the [[Shoelace Theorem]], the area of <math>\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12)|=|\dfrac{1}{2}(-10)|=5</math>.
  
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==Solution 4==
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This means the fraction of the total area is <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}</math>.
 +
 
 +
===Solution 4===
  
 
The smallest rectangle that follows the grid lines and completely encloses <math>\triangle ABC</math> has an area of <math>12</math>, where <math>\triangle ABC</math> splits the rectangle into four triangles. The area of <math>\triangle ABC</math> is therefore <math>12 - (\frac{4 \cdot 2}{2}+\frac{3 \cdot 1}{2}+\frac{3 \cdot 1}{2}) = 12 - (4 + \frac{3}{2} + \frac{3}{2}) = 12 - 7 = 5</math>. That means that <math>\triangle ABC</math> takes up <math>\frac{5}{30} = \boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid.
 
The smallest rectangle that follows the grid lines and completely encloses <math>\triangle ABC</math> has an area of <math>12</math>, where <math>\triangle ABC</math> splits the rectangle into four triangles. The area of <math>\triangle ABC</math> is therefore <math>12 - (\frac{4 \cdot 2}{2}+\frac{3 \cdot 1}{2}+\frac{3 \cdot 1}{2}) = 12 - (4 + \frac{3}{2} + \frac{3}{2}) = 12 - 7 = 5</math>. That means that <math>\triangle ABC</math> takes up <math>\frac{5}{30} = \boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid.
  
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==Solution 5==
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===Solution 5 (Very much recommended to learn this)===
 +
 
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Using [[Pick's Theorem]], the area of the triangle is <math>4 + \dfrac{4}{2} - 1=5</math>. Therefore, the triangle takes up <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid.
 +
 
 +
===Solution 6 (Heron's Formula, Not Recommended)===
 +
 
 +
We can find the lengths of the sides by using the [[Pythagorean Theorem]]. Then, we apply [[Heron's Formula]] to find the area.
 +
<cmath> \sqrt{\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-2\sqrt{5}\right)}. </cmath>
 +
This simplifies to
 +
<cmath> \sqrt{\left(\sqrt{10}+\sqrt{5}\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{10}\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{10}\right)\left(\sqrt{10}+\sqrt{5}-2\sqrt{5}\right)}. </cmath>
 +
Again, we simplify to get
 +
<cmath> \sqrt{\left(\sqrt{10}+\sqrt{5}\right)\left(\sqrt{5}\right)\left(\sqrt{5}\right)\left(\sqrt{10}-\sqrt{5}\right)}. </cmath>
 +
The middle two terms inside the square root multiply to <math> 5 </math>, and the first and last terms inside the square root multiply to <math> \sqrt{10}^2-\sqrt{5}^2=10-5=5. </math> This means that the area of the triangle is
 +
<cmath> \sqrt{5\cdot 5}=5. </cmath>
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The area of the grid is <math> 6\cdot 5=30. </math> Thus, the answer is <math> \frac{5}{30}=\boxed{\textbf{(A) }\frac{1}{6}} </math>.
 +
===Solution 7 (Simple Deduction)===
 +
First, count the number of shapes inside the main triangle (you should count 10). Then, upon closer inspection, most of the shapes that are not a single unit on the triangle can be created by connecting another shape. The only exceptions are one shape that is a single unit and one that would need 2 shapes connected to it to make a single unit. On average, you need to connect 2 shapes to make a unit. Knowing this, if there are 10 shapes and you require 2 shapes to make a unit, 10 divided by 2 equals 5, which is the area.5 is 1/6 of 30(the total of the graph) and so the final answer is <math> \frac{5}{30}=\boxed{\textbf{(A) }\frac{1}{6}} </math>.
 +
-Themathnerd3.14
 +
 
 +
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/ZOEG-KfBA2E
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
==Video Solution==
 +
https://youtu.be/EyDGtLc6xGE
 +
 
 +
~savannahsolver
  
−
Using Pick's Theorem, the area of the triangle is <math>4 + \dfrac{4}{2} - 1=5</math>. Therefore, the triangle takes up <math>\dfrac{5}{30}=\boxed{\textbf{(A)}~\frac{1}{6}}</math> of the grid.
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==Video Solution==
 +
https://youtu.be/j3QSD5eDpzU?t=507
  
−
Solution by bobert1
 
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2015|num-b=18|num-a=20}}
 
{{AMC8 box|year=2015|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:22, 11 July 2024

Problem

A triangle with vertices as $A=(1,3)$, $B=(5,1)$, and $C=(4,4)$ is plotted on a $6\times5$ grid. What fraction of the grid is covered by the triangle?

$\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}$

[asy] draw((1,0)--(1,5),linewidth(.5)); draw((2,0)--(2,5),linewidth(.5)); draw((3,0)--(3,5),linewidth(.5)); draw((4,0)--(4,5),linewidth(.5)); draw((5,0)--(5,5),linewidth(.5)); draw((6,0)--(6,5),linewidth(.5)); draw((0,1)--(6,1),linewidth(.5)); draw((0,2)--(6,2),linewidth(.5)); draw((0,3)--(6,3),linewidth(.5)); draw((0,4)--(6,4),linewidth(.5)); draw((0,5)--(6,5),linewidth(.5)); draw((0,0)--(0,6),EndArrow); draw((0,0)--(7,0),EndArrow); draw((1,3)--(4,4)--(5,1)--cycle); label("$y$",(0,6),W); label("$x$",(7,0),S); label("$A$",(1,3),dir(210)); label("$B$",(5,1),SE); label("$C$",(4,4),dir(100)); [/asy]

Solutions

Solution 1

The area of $\triangle ABC$ is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is $\sqrt{1^2+2^2}=\sqrt{5}$, and its base is $\sqrt{2^2+4^2}=\sqrt{20}$. We multiply these and divide by $2$ to find the area of the triangle is $\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5$. Since the grid has an area of $30$, the fraction of the grid covered by the triangle is $\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}$.

Solution 2

Note angle $\angle ACB$ is right; thus, the area is $\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=10 \times \dfrac{1}{2}=5$; thus, the fraction of the total is $\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}$.

Solution 3

By the Shoelace Theorem, the area of $\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12)|=|\dfrac{1}{2}(-10)|=5$.

This means the fraction of the total area is $\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}$.

Solution 4

The smallest rectangle that follows the grid lines and completely encloses $\triangle ABC$ has an area of $12$, where $\triangle ABC$ splits the rectangle into four triangles. The area of $\triangle ABC$ is therefore $12 - (\frac{4 \cdot 2}{2}+\frac{3 \cdot 1}{2}+\frac{3 \cdot 1}{2}) = 12 - (4 + \frac{3}{2} + \frac{3}{2}) = 12 - 7 = 5$. That means that $\triangle ABC$ takes up $\frac{5}{30} = \boxed{\textbf{(A)}~\frac{1}{6}}$ of the grid.

Solution 5 (Very much recommended to learn this)

Using Pick's Theorem, the area of the triangle is $4 + \dfrac{4}{2} - 1=5$. Therefore, the triangle takes up $\dfrac{5}{30}=\boxed{\textbf{(A)}~\frac{1}{6}}$ of the grid.

Solution 6 (Heron's Formula, Not Recommended)

We can find the lengths of the sides by using the Pythagorean Theorem. Then, we apply Heron's Formula to find the area. \[\sqrt{\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-\sqrt{10}\right)\left(\frac{\sqrt{10}+\sqrt{10}+2\sqrt{5}}{2}-2\sqrt{5}\right)}.\] This simplifies to \[\sqrt{\left(\sqrt{10}+\sqrt{5}\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{10}\right)\left(\sqrt{10}+\sqrt{5}-\sqrt{10}\right)\left(\sqrt{10}+\sqrt{5}-2\sqrt{5}\right)}.\] Again, we simplify to get \[\sqrt{\left(\sqrt{10}+\sqrt{5}\right)\left(\sqrt{5}\right)\left(\sqrt{5}\right)\left(\sqrt{10}-\sqrt{5}\right)}.\] The middle two terms inside the square root multiply to $5$, and the first and last terms inside the square root multiply to $\sqrt{10}^2-\sqrt{5}^2=10-5=5.$ This means that the area of the triangle is \[\sqrt{5\cdot 5}=5.\] The area of the grid is $6\cdot 5=30.$ Thus, the answer is $\frac{5}{30}=\boxed{\textbf{(A) }\frac{1}{6}}$.

Solution 7 (Simple Deduction)

First, count the number of shapes inside the main triangle (you should count 10). Then, upon closer inspection, most of the shapes that are not a single unit on the triangle can be created by connecting another shape. The only exceptions are one shape that is a single unit and one that would need 2 shapes connected to it to make a single unit. On average, you need to connect 2 shapes to make a unit. Knowing this, if there are 10 shapes and you require 2 shapes to make a unit, 10 divided by 2 equals 5, which is the area.5 is 1/6 of 30(the total of the graph) and so the final answer is $\frac{5}{30}=\boxed{\textbf{(A) }\frac{1}{6}}$. -Themathnerd3.14

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/ZOEG-KfBA2E

~Education, the Study of Everything


Video Solution

https://youtu.be/EyDGtLc6xGE

~savannahsolver

Video Solution

https://youtu.be/j3QSD5eDpzU?t=507

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
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Problem 18 		Followed by
Problem 20
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