AMC 8 2015 Problem 19

AMC8 2015 年第 19 题

坐标几何★★★☆☆

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?

三个顶点为 A＝(1,3)，B＝(5,1)，C＝（4,4）的三角形画在一张 6ｘ5 的方格纸上，这张方格纸上被三角形覆盖的面积所占总面积的比例是多少？

A. $\displaystyle \frac{1}{6}$
B. $\displaystyle \frac{1}{5}$
C. $\displaystyle \frac{1}{4}$
D. $\displaystyle \frac{1}{3}$
E. $\displaystyle \frac{1}{2}$

正确答案 · Correct Answer

解析 · Solution

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

\displaystyle \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

\displaystyle \frac{5}{30} = \frac{1}{6}

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

\displaystyle 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

\displaystyle \frac{5}{30} = \frac{1}{6}

of the grid. Using Pick's Theorem, the area of the triangle is

4 + \dfrac{4}{2} - 1=5

. Therefore, the triangle takes up

\dfrac{5}{30}= \frac{1}{6}

of the grid. We can find the lengths of the sides by using the Pythagorean Theorem. Then, we apply Heron's Formula to find the area.

\displaystyle \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

\displaystyle \frac{5}{30} = \frac{1}{6}

. First, count the number of 1x1 squares inside the main triangle, both fully enclosed and partially (you should count 10). Then, upon closer inspection, most of the squares that are not a single unit on the triangle can be created by connecting another shape. The only exception is one square 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 squares to make a unit. Knowing this, if there are 10 squares and you require 2 shapes to make a unit, 10 divided by 2 equals 5, which is the area.5 is

\displaystyle \frac{1}{6}

of 30(the total of the graph) and so the final answer is

\displaystyle \frac{5}{30} = \frac{1}{6}

. -Themathnerd3.14

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