Triangle mother

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Gauss's Shoelace Formula

Mid Point of Line Segment

Equation of The Line

I think that my solution isn't the most simplest one, but I enjoyed so much solving this way!

The Gauss's Shoelace Formula allows us to find a area of a simple polygon whose vertices are in the Cartesian Plane by using matrices. Observe the following image:

We multiply two numbers connected by a red slash and sum up the products. We obtain $19p+240+23q$ . Then we multiply two numbers connected by a green slash and sum up the products. We obtain $12q+437+20p$ .

The modulus of the difference between these two results is 2 times the area of Triangle $ABC$ which is $2\times70=140$ . Then we have $|(19p+240+23q)-(12q+437+20p)|=|p+197-11q|=140$ .

Now we can get another equation by using the information about the median.

The median of a triangle is the segment that connects a vertex to the midpoint of its opposite side. Then we need to find the midpoint of the segment $BC$ . The coordinates of the segment that connects points $(x_1,y_1)$ and $(x_2,y_2)$ is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .

Then, the coordinates of the midpoint $M$ of the segment that connects points $B(12,19)$ and $C(23,20)$ is $\left(\frac{12+23}{2},\frac{19+20}{2}\right)=M(17.5,19.5)$

Now we need to find the equation of the line that passes trough points $M(17.5,19.5)$ and $A(p, q)$ . To do so I will use my favorite method that use the following formula to find the general equation of a line. This formula can be obtained by using similar triangles:

Equation of the line that passes trough points $(x_1,y_1)$ and $(x_2,y_2)$ is $\left(\frac{x-x_1}{x_1-x_2}=\frac{y-y_1}{y_1-y_2}\right)$ . Solving for points $A(p, q)$ and $M(17.5,19.5)$ :

$\left(\frac{x-p}{p-17.5}=\frac{y-q}{q-19.5}\right)$

$(x-p)(q-19.5)=(y-q)(9-17.5)$

$xq-19.5x-pq+19.5p=yp-17.5y-pq+17.5q$

$xq-19.5x+19.5p=yp-17.5y+17.5q$

Manipulation the equation to get the reduced equation in the form $ma+b=y$ where $m$ is the slope of the line:

$xq-19,5x+19,5p=yp-17,5y+17,5q$

$xq-19.5x+19.5p-17.5q=yp-17.5y$

$x(q-19.5)+19.5p-17.5q=y(p-17.5)$

$\frac{(q-19.5)}{(p-17.5)}x+\frac{(19.5p-17.5q)}{(p-17.5)}=y$

We can see in this equation by compared to the reduced one that our $m=\frac{(q-19.5)}{(p-17.5)}$ . In this case the slope is equal to $-5$ . Then we have $\frac{(q-19.5)}{(p-17.5)}=-5$

Now we have the following system of equations:

$\begin{cases} \frac { q-19.5 }{ p-17.5 }=-5 \\ |p+197-11q|=140 \end{cases}$

For the second equation we have two cases: (i) $p+197-11q=140$ and (ii) $p+197-11q=-140$ . Solving for each case:

Case (i)

$p+197-11q=140$

$p=11q-57$

Substituting in the first equation:

$\frac { q-19.5 }{ p-17.5 }=-5$

$\frac { q-19.5}{ 11q-57-17.5 }=-5$

$\frac { q-19.5 }{ 11q-74,5}=-5$

$q-19.5 =-5(11q-74,5)$

$q-19.5 =(-55q+372,5)$

$56q =(392$

$q=7$

$p=11\cdot(7)-57=20$

Then we have $p=20$ and $q=7$ . Then $p+q=20+7=27$ in this case.

Case (ii)

$p+197-11q=-140$

$p=11q-377$

Substituting in the first equation:

$\frac { q-19.5 }{ p-17.5 }=-5$

$\frac { q-19.5}{ 11q-337-17.5 }=-5$

$\frac { q-19.5 }{ 11q-354,5}=-5$

$q-19.5 =-5(11q-354,5)$

$q-19.5 =(-55q+1792)$

$56q =(1792$

$q=32$

$p=11\cdot(32)-57=15$

Then we have $p=15$ and $q=32$ .

Then $p+q$ will be $15+32=47$ in this case.

$47>27$

Hence the largest possible value for $p+q$ is $\boxed{47}$