Finding the area of an orthic triangle (DEF) when given vertices of triangle ABC.

Question

Given the triangle ABC whose vertices are endpoints of the altitudes from $A$,$B$ and $C$ is called the orthic triangle. The triangle $ABC$ has vertices at $A=(2,4)$; $B=(8,5)$ and $C=(3,9)$. The altitude from $B$ to $AC$, meets AC at point $D=(2.42,6.12)$. Find the area if the orthic triangle.

To attempt this problem I decided to use the formula $$area = \frac{abc|cosAcosBcosC|}{2R}$$ where $R$ is the circumradius of the triangle $ABC$. I calculated the length of each sides by using the length equation and got $a=\sqrt41$, $b=\sqrt26$, and $c=\sqrt37$.

Next I used the cosine rule to get the angle at vertix $A$. Then used the sine rule to get angle at $B$ and subtracted these 2 angles from 180 to get the third angle. $A=69.23$, $B=48.12$ and $C=62.65$.

To get the circumradius of the triangle $ABC$ I used $$R=\frac{abc}{4(Area)}$$ I found the area using heron's formula to be 14.5, putting all these values into the original equation I got the area of the orthic triangle to be 3.154. Is this correct? Is there an easier method? Thanks

Answer 1

I suggest to use $$ \Delta_{orthic} = 2\Delta\left|\cos A\cos B\cos C\right|$$ and to compute $\Delta$ from the shoelace formula and the remaining part from the cosine theorem, such that we do not need to extract any square root. By the shoelace formula

$$ 2\Delta = |2\cdot 5+8\cdot 9+3\cdot 4-4\cdot 8-5\cdot 3-9\cdot 2|=29 $$ and by the cosine theorem $$\left|\cos A\cos B\cos C\right|=\frac{(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^2+b^2+c^2)}{8a^2b^2 c^2} $$ where $a^2=41,b^2=26,c^2=37$ are given by the Pythagorean theorem. It follows that $$\Delta_{orthic} = 29\cdot\frac{30\cdot 52\cdot 22 }{8\cdot 41\cdot 26\cdot 37}=\frac{4785}{1517}. $$

Answer 2

Find the feet of the altitudes $D$, $E$, $F$ (you already found $D$). The area of $DEF$ is $\displaystyle \frac12 \left\Vert \overrightarrow{DE}\times\overrightarrow{DF}\right\Vert$.

To find $E$, we first find the equation of $(BC)$: $\displaystyle y=\dfrac{9-5}{3-8}(x-3)+9=-4/5x+11.4$. The equation of the altitude $(AE)$ is $\displaystyle y=5/4 (x-2)+4=5/4 x+3/2$. To find the intersection $E$ of these two lines we solve $\displaystyle 5/4 x+3/2=-4/5x +11.4$ we get $\displaystyle E\left(\frac{198}{41},\frac{309}{41}\right)$.

After you find $D$, $E$ and $F$ you can use the formula I gave above. If you don't know what's a cross product, you can use the answers in this question.