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2015 AMC 8 Problems/Problem 19: Difference between revisions

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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>
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));
Line 23: Line 23:
</asy>
</asy>


==Solution 1==
==Solutions==


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>.
===Solution 1===


==Solution 2==
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>


==Solution 3==
===Solution 2===
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>.
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>.


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


==Solution 4==
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.


==Solution 5==
===Solution 5 (Very much recommended to learn this)===


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.
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)==
===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.  
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>
<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
This simplifies to
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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
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>
<cmath> \sqrt{5\cdot 5}=5. </cmath>
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>. -BorealBear
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
 
==Video Solution by OmegaLearn==
https://youtu.be/j3QSD5eDpzU?t=507


==See Also==
==See Also==
Line 59: Line 79:
{{AMC8 box|year=2015|num-b=18|num-a=20}}
{{AMC8 box|year=2015|num-b=18|num-a=20}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Introductory Algebra Problems]]

Latest revision as of 18:33, 28 June 2025

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 by OmegaLearn

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

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 AJHSME/AMC 8 Problems and Solutions

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