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

## Problem

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

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

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

## 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 $4\times 3=12$

## Solution 4

Pick's theorem states $A = I + \frac{1}{2}B - 1$, where B is the number of boundary points and I is the number of interior points. There are 3 boundary points and 0 interior points. Substituting the values in, we get that the answer is $\frac{1}{2}$. ~ math_genius_11