Obtuse isosceles orthic triangles – High School Geometry

In a non-right triangle, each of the following six statements

(1) $\begin{equation*} \begin{split} h_C&=R\cos C\\ \cos C&=\frac{2ab}{a^2+b^2}\\ a^2+b^2&=4R^2\\ h_A^2+h_B^2&=AB^2\\ (a^2-b^2)^2&=(ac)^2+(cb)^2\\ a\cos A+b\cos B &=0\\ \end{split} \end{equation*}$

implies the others. Further, under this equivalence, the orthic triangle of the parent triangle is necessarily isosceles.

Obvious things

Observe that $a\neq b$ , and so a triangle that satisfies the entire chain of equivalence in (1) cannot be equilateral. Also, a right triangle with hypotenuse $c$ satisfies both $a^2+b^2=4R^2$ and $h_A^2+h_B^2=AB^2$ , but not the rest. Hence, the non-right triangle requirement is essential.

Obtuse triangle

As the post’s title gives away, the first obvious thing to get out of the way.

If the equivalence in (1) holds, PROVE that the parent triangle contains an obtuse angle.

This is evident from $a\cos A+b\cos B=0$ .

Other techniques

Alternatively, one can compute the cosine of $\angle A$ or $\angle B$ explicitly, using $(a^2-b^2)^2=(ac)^2+(cb)^2$ and the cosine formula:

$\begin{equation*} \begin{split} \cos A&=\frac{b^2+c^2-a^2}{2bc}\\ &=\frac{b^2+\left(\frac{(a^2-b^2)^2}{a^2+b^2}\right)-a^2}{2bc}\\ &=\frac{b(b^2-a^2)}{c(a^2+b^2)} \end{split} \end{equation*}$

Similarly, $\cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}$ . Now $a\neq b$ implies either $a< b$ or $b< a$ . In turn, either $\cos A< 0$ or $\cos B< 0$ , showing that one of the two angles $\angle A,\angle B$ is obtuse.

Oversimple theorem

Now, let’s attempt to make our simple equivalence even simpler by splitting the proof into six parts.

If $h_C=R\cos C$

, PROVE that $\cos C=\frac{2ab}{a^2+b^2}$

From $h_C=R\cos C$ , isolate $\cos C$ and then use the fact that $h_C=\frac{ab}{2R}$ :

$\begin{equation*} \begin{split} \cos C&=\frac{h_C}{R}\\ &=\frac{ab}{2R^2}\\ &=\frac{ab}{2\left(\frac{c}{2\sin C}\right)^2}\\ c^2\cos C&=2ab\sin^2 C\\ (a^2+b^2-2ab\cos C)\cos C&=2ab(1-\cos^2 C)\\ \implies \cos C&=\frac{2ab}{a^2+b^2} \end{split} \end{equation*}$

If $\cos C=\frac{2ab}{a^2+b^2}$

, PROVE that $a^2+b^2=4R^2$

By the extended law of sines, we have $R=\frac{c}{2\sin C}$ . So:

$\begin{equation*} \begin{split} 4R^2&=4\left(\frac{c}{2\sin C}\right)^2\\ &=\frac{c^2}{\sin^2 C}\\ &=\frac{c^2}{1-\cos^2 C}\\ &=\frac{a^2+b^2-2ab\cos C}{1-\cos^2 C}\\ &=\frac{a^2+b^2-2ab\left(\frac{2ab}{a^2+b^2}\right)}{1-\left(\frac{2ab}{a^2+b^2}\right)^2}\\ &=\frac{\frac{(a^2+b^2)^2-4a^2b^2}{a^2+b^2}}{\frac{(a^2+b^2)^2-4a^2b^2}{(a^2+b^2)^2}}\\ &=a^2+b^2 \end{split} \end{equation*}$

If $a^2+b^2=4R^2$

, PROVE that $h_A^2+h_B^2=AB^2$

You’ll notice the semblance with our previous post. Repeated for emphasis.

$\begin{equation*} \begin{split} h_A^2+h_B^2&=\left(\frac{bc}{2R}\right)^2+\left(\frac{ac}{2R}\right)^2\\ &=\left(\frac{c}{2R}\right)^2(b^2+a^2)\\ &=\left(\frac{c}{2R}\right)^2(4R^2)\\ &=c^2=AB^2~~\textrm{as per usual notation} \end{split} \end{equation*}$

If $h_A^2+h_B^2=AB^2$

, PROVE that $(a^2-b^2)^2=(ac)^2+(cb)^2$

From $AB^2=h_A^2+h_B^2$ , we have:

$\begin{equation*} \begin{split} c^2&=\left(\frac{bc}{2R}\right)^2+\left(\frac{ac}{2R}\right)^2\\ \implies a^2+b^2-2ab\cos C&=\frac{c^2}{4R^2}(a^2+b^2)\\ 4R^2(a^2+b^2-2ab\cos C)&=c^2(a^2+b^2)\\ 4\left(\frac{c}{2\sin C}\right)^2(a^2+b^2-2ab\cos C)&=c^2(a^2+b^2)\\ c^2(a^2+b^2-2ab\cos C)&=c^2(a^2+b^2)(1-\cos^2 C)\\ \implies \cos C&=\frac{2ab}{a^2+b^2},0\\ \end{split} \end{equation*}$

Since we’re in a non-right triangle setting, we choose $\cos C=\frac{2ab}{a^2+b^2}$ . Then compute $c^2$ :

$c^2=a^2+b^2-2ab\left(\frac{2ab}{a^2+b^2}\right)\implies (a^2-b^2)^2=(ac)^2+(cb)^2.$

If $(a^2-b^2)^2=(ac)^2+(cb)^2$

, PROVE that $a\cos A+b\cos B=0$

From $(a^2-b^2)^2=(ac)^2+(cb)^2$ , isolate $c^2=\frac{(a^2-b^2)^2}{a^2+b^2}$ . Then:

$\begin{equation*} \begin{split} a\cos A+b\cos B&=a\left(\frac{b^2+c^2-a^2}{2bc}\right)+b\left(\frac{a^2+c^2-b^2}{2ac}\right)\\ &=\scriptstyle a\left(\frac{b^2+\left(\frac{(a^2-b^2)^2}{a^2+b^2}\right)-a^2}{2bc}\right)+b\left(\frac{a^2+\left(\frac{(a^2-b^2)^2}{a^2+b^2}\right)-b^2}{2ac}\right)\\ &=\frac{ab(b^2-a^2)}{c(a^2+b^2)}+\frac{ab(a^2-b^2)}{c(a^2+b^2)}\\ &=0 \end{split} \end{equation*}$

If $a\cos A+b\cos B=0$

, PROVE that $h_C=R\cos C$

Note that $a\cos A+b\cos B=0$ gives $\cos C=\frac{2ab}{a^2+b^2}$ . So:

$\begin{equation*} \begin{split} R\cos C&=R\left(\frac{2ab}{a^2+b^2}\right)\\ h_C&=\frac{ab}{2R}\\ &=R\left(\frac{2ab}{4R^2}\right)\\ &=R\left(\frac{2ab}{4\left(\frac{c}{2\sin C}\right)^2}\right)\\ &=R\left(2ab\times\frac{\sin^2 C}{c^2}\right)\\ &=R\left(2ab\times\frac{1-\left(\frac{2ab}{a^2+b^2}\right)^2}{a^2+b^2-2ab\left(\frac{2ab}{a^2+b^2}\right)}\right)\\ &=R\left(\frac{2ab}{a^2+b^2}\right)\\ \therefore h_C&=R\cos C. \end{split} \end{equation*}$

Six. Perfect.

Orthic triangle

The triangle obtained by joining the feet of the three altitudes of a parent triangle is called its orthic triangle.

Suppose that a non-right triangle satisfies equivalence (1). PROVE that its orthic triangle is necessarily isosceles.

The lengths of the sides of an orthic triangle are given by $a\cos A$ , $b\cos B$ , $c\cos C$ (one of these may require an absolute value if the parent triangle is obtuse). One of our equivalent statements is $a\cos A=-b\cos B$ . Thus, the orthic triangle has two sides equal. And we can even make it three.

Suppose that a triangle satisfies equivalence (1). PROVE the extra equivalence: $a^2=3b^2\iff$

the parent triangle is isosceles with $b=c\iff$

the orthic triangle is equilateral.

First suppose that $a^2=3b^2$ . Using the cosine formula:

$\begin{equation*} \begin{split} c^2&=a^2+b^2-2ab\cos C\\ &=3b^2+b^2-2(\sqrt{3}b)b\left(\frac{2\sqrt{3}b\times b}{3b^2+b^2}\right)\\ &=b^2 \end{split} \end{equation*}$

Thus, $c=b$ . Next, suppose that the parent triangle is isosceles with $b=c$ . We’ll show that its orthic triangle is equilateral. Indeed, we already had $a\cos A=-b\cos B$ . It now suffices to show that $c\cos C=b\cos B$ . By the cosine formula:

$\begin{equation*} \begin{split} c\cos C&=c\left(\frac{a^2+b^2-c^2}{2ab}\right)\\ &=c\left(\frac{a^2}{2ab}\right)\\ &=\frac{a}{2}\\ b\cos B&=b\left(\frac{a^2+c^2-b^2}{2ac}\right)\\ &=b\left(\frac{a^2}{2ac}\right)\\ &=\frac{a}{2}\\ \end{split} \end{equation*}$

Thus, the orthic triangle is equilateral. Finally, suppose that the orthic triangle is equilateral. Consider $b\cos B=c\cos C$ :

$\begin{equation*} \begin{split} b\left(\frac{a^2+c^2-b^2}{2ac}\right) &=c\left(\frac{a^2+b^2-c^2}{2ab}\right)\\ (c^2)^2-a^2c^2+b^2(a^2-b^2)&=0\\ \implies c^2&=\frac{a^2\pm\sqrt{(-a^2)^2-4b^2(a^2-b^2)}}{2}\\ c^2&=\frac{a^2\pm(a^2-2b^2)}{2}\\ \therefore c^2&=a^2-b^2,\quad\textrm{or}\quad c^2=b^2\\ \end{split} \end{equation*}$

Discard $c^2=a^2-b^2$ since we’re dealing with a non-right triangle. The second possibility then yields $c=b$ . Now use $c=b$ in the equation $(a^2-b^2)^2=(ac)^2+(cb)^2$ :

$(a^2-b^2)^2=b^2(a^2+b^2)\implies a^2(a^2-3b^2)=0\implies a^2=3b^2.$

All this trouble just for a single triangle: the one in which $\angle A=120^{\circ}$ , $\angle B=30^{\circ}$ , $\angle C=30^{\circ}$ . That’s the triangle that satisfies the equivalent conditions just established.

Consider triangle $ABC$

with vertices at $A(-6,0)$

, $B(0,0)$

, and $C(2,4)$

. Verify that its orthic triangle is isosceles.

The feet of the altitudes from $A,B,C$ are $F_A(-1.2,-2.4)$ , $F_B(-1.2,2.4)$ , and $F_C(2,0)$ , as shown above. By the distance formula:

$F_AF_B=4.8,~ F_BF_C=4,~F_CF_A=4.$

Notice that the foot of the altitude from $C$ coincides with our favourite point $W$ .

Takeaway

For any non-right triangle $ABC$ satisfying the chain of equivalence in (1), the three statements below are equivalent:

$b=c$
$a^2=3b^2$
the orthic triangle is equilateral.

Thus, there’s only “one” strictly isosceles triangle whose orthic triangle is equilateral.

Tasks

Let be the interior angles of a triangle , and let be the side-lengths. Relative to these measurements, PROVE that:
- the interior angles of the orthic triangle are $180^{\circ} - 2 \angle A$ , $180^{\circ} - 2 \angle B$ , $180^{\circ} - 2 \angle C$ , if the parent triangle is acute
- the interior angles of the orthic triangle are $2 \angle A-180^{\circ}$ , $2 \angle B$ , $2 \angle C$ , if the parent triangle contains an obtuse angle $\angle A$
- the side-lengths of the orthic triangle are $a\cos A$ , $b\cos B$ , $c\cos C$ (one of these will be negative if the parent triangle contains an obtuse angle, in which case an absolute value will be needed).
Let be a real number. Consider with interior angles , , and .
- What restrictions, if any, must be imposed on $\theta$ ?
- PROVE that the orthic triangle associated with $\triangle ABC$ is isosceles.
PROVE that the two statements below are equivalent for a triangle :
- $a^2=3b^2$ and $b=c$
- $\angle A=120^{\circ}$ , $\angle B=30^{\circ}$ , $\angle C=30^{\circ}$ .
Consider with vertices at , , . Find:
- the coordinates of the feet of the three altitudes
- the lengths of the sides of the orthic triangle
- a condition for which the orthic triangle will be equilateral.
PROVE that the following statements are equivalent for any triangle:
- the orthic triangle is equilateral
- the parent triangle is equilateral or the parent triangle is isosceles with base angles of $30^{\circ}$ and an apex angle of $120^{\circ}$ .