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2015 AMC 8 Problems/Problem 19

Revision as of 23:07, 16 May 2023 by Thestudyofeverything (talk | contribs) (→‎Video Solution by OmegaLearn)

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

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 $\frac{5}{30}=\boxed{\textbf{(A) }\frac{1}{6}}$.

-BorealBear

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)
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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