Difference between revisions of "1996 AJHSME Problems/Problem 22"

Revision as of 23:20, 15 August 2020

Problem

The horizontal and vertical distances between adjacent points equal 1 unit. The area of triangle $ABC$ is

$[asy] for (int a = 0; a < 5; ++a) { for (int b = 0; b < 4; ++b) { dot((a,b)); } } draw((0,0)--(3,2)--(4,3)--cycle); label("$A$",(0,0),SW); label("$B$",(3,2),SE); label("$C$",(4,3),NE); [/asy]$

$\text{(A)}\ 1/4 \qquad \text{(B)}\ 1/2 \qquad \text{(C)}\ 3/4 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 5/4$

Solution 1

$[asy] for (int a = 0; a < 5; ++a) { for (int b = 0; b < 4; ++b) { dot((a,b)); } } draw((0,0)--(3,2)--(4,3)--cycle); draw((0,0)--(3,2)--(4,0)--cycle); draw((4,2)--(3,2)--(4,3)--cycle); draw((0,0)--(4,0)--(4,3)--cycle); draw((3,2)--(3,0)--cycle); label("$A$",(0,0),SW); label("$B$",(3,2),SE); label("$C$",(4,3),NE); label("$D$",(4,0),SE); label("$E$",(4,2),SE); label("$F$",(3,0),SE); [/asy]$

$\triangle ADC$ takes up half of the 4x3 grid, so it has area of $6$ .

$\triangle ABD$ has height of $BF = 2$ and a base of $AD = 4$ , for an area of $\frac{1}{2}\cdot 2 \cdot 4 = 4$ .

$\triangle CBD$ has height of $BE = 1$ and a base of $CD = 3$ , for an area of $\frac{1}{2}\cdot 1 \cdot 3 = \frac{3}{2}$

Note that $\triangle ABC$ can be found by taking $\triangle ADC$ , and subtracting off $\triangle ABD$ and $\triangle CBD$ .

Thus, the area of $\triangle ABC = 6 - 4 - \frac{3}{2} = \frac{1}{2}$ , and the answer is $\boxed{B}$ .

There are other equivalent ways of dissecting the figure; right triangles $\triangle ABF, \triangle BCE$ and rectangle $\square BEDF$ can also be used. You can also use $\triangle{BEC}$ and trapezoid $ADBE$ .

Solution 2

Using the Shoelace Theorem, and labelling the points $A(0,0), B(3,2), C(4,3)$ , we find the area is:

$(0,0)$

$(3,2)$

$(4,3)$

$(0,0)$

Area = $\frac{1}{2} = |(3\cdot 0 + 4\cdot 2 + 0\cdot 3) - (0\cdot 2 + 3\cdot 3 + 4\cdot 0)| = \frac{1}{2}$ , which is option $\boxed{B}$ .

Solution 3

We can find the total area of the grid, and subtract the triangles off of the area. The area of the grid is $4x3=12$

Solution 4

Pick's theorem says $A = I + \frac{1}{2}B - 1$ , where B is the # of boundary points and I is the # of interior points. There are 3 boundary points and 0 interior points. Substituting the values in, we get that the answer is $\frac{1}{2}$

@@ Line 71: / Line 71: @@
 ==Solution 3==
 We can find the total area of the grid, and subtract the triangles off of the area. The area of the grid is <math>4x3=12</math>
+==Solution 4==
+Pick's theorem says  <math>A = I + \frac{1}{2}B - 1</math>, where B is the # of boundary points and I is the # of interior points. There are 3 boundary points and 0 interior points. Substituting the values in, we get that the answer is <math>\frac{1}{2}</math>
 ==See Also==

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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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