# What is the minimum area of triangle [closed]

In triangle $ABC$, the respective coordinates of $A$ and $B$ are $(0,0)$ and $(15,20)$. It is known that $C$ has integer coordinates. What is the minimum positive area of triangle $ABC$?

Really, I have no idea what to do.

## closed as off-topic by Trevor Gunn, I am Back, Henrik, user91500, Especially LimeJun 8 '17 at 9:50

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• There is a formula using a determinant to compute the area of a given triangle. For example, see math.stackexchange.com/questions/299352/… – Zuriel Jun 8 '17 at 3:01
• en.wikipedia.org/wiki/Pick%27s_theorem – Jack D'Aurizio Jun 8 '17 at 3:02
• Please type in your question instead of posting it as an image. An image takes longer to download, is not searchable, and cannot be seen on some devices. – Joel Reyes Noche Jun 8 '17 at 3:20
• Although it's a pity the OP didn't show any sign of having done his own research, it's very pleasant to see these 3 different, quite elegant answers, which all came very quickly. – Joce Jun 8 '17 at 7:18

Let the third point of triangle be given by $(h,k)$.

Thus the area of triangle formed by : $(0,0) ,(15,20)$ & $(h,k)$ is -

\begin{align} \text{area} ~&=\left |\frac 12 \det \begin{pmatrix}0 & 0 & 1\\ h & k & 1 \\ 15 & 20 & 1 \\ \end{pmatrix} \right |\\ &=\frac{5}{2} \left |4h-3k \right | \end{align}

Now you need to minimize $4h-3k$ in order to get least positive area. Since $h$ and $k$ are integers, you can easily conclude that minimum positive value of $4h-3k$ can be $1$ (For example, at $h=1$ and $k=1$). Hence, the desired area is $\dfrac{5}{2}$.

Using only geometry and $\text{Area}=\frac{1}{2} \text{base} \times \text{height}$.

Consider $AB$ as the base, minimize the height.

Consider the lines $$y=\frac{4}{3}x+c$$

Find the minimum value of $|c|$ so that there are integer solutions.

So that's equivalent to finding the minimum positive value of $|3y-4x|$ for integers $x$, $y$.

Since $x,y$ are integers, so the minimum possible value is $|3y-4x|$ is $1$, which occurs when e.g., $x=y=1$. So the minimum value of $|c|$ is $1/3$.

The area of the triangle is $$\text{Area}=\frac{1}{2}\sqrt{15^2+20^2}\left(\frac{1}{3}\frac{15}{\sqrt{15^2+20^2}}\right)=\frac{5}{2}$$

$\newcommand{\vec}[1]{\boldsymbol{#1}}$A triangle spanned by vectors $\vec{u},\vec{v}$ has area $$S = \frac12 |\vec{u} \times \vec{v}|$$ so we can set $\vec{u} := [15,20]^\intercal,\vec{v} = [x,y]^\intercal$ with $x,y$ being integers, to be the two edges of the triangle.

The area is then $$S= \frac12 \left|15y - 20x\right|$$ Since $\gcd(15,20) = 5$, we can expect $\frac52$ to be the minimal (positive) value of $S$, which occurs at $x = 2,y = 3$.