General – Page 20 – High School Geometry

Obtuse isosceles orthic triangles

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 $120^{\circ}$ .

A note on Kepler triangle

A Kepler triangle is first of all a right triangle, and so it enjoys the Pythagorean identity:

(1) $\begin{equation*} c^2=a^2+b^2 \end{equation*}$

In addition, it has the property that its side-lengths form a geometric progression, for example:

(2) $\begin{equation*} b,a,c\iff a^2=bc \end{equation*}$

Equations (1) and (2) force the ratio of the hypotenuse $c$ to one of the legs ( $a$ or $b$ ) to be golden:

(3) $\begin{equation*} \frac{c}{b}=\varphi,~~\varphi:=\textrm{golden ratio} \end{equation*}$

In this post, we show that a Kepler triangle also satisfies

(4) $\begin{equation*} (ac)^2=(ab)^2+(bc)^2 \end{equation*}$

which, in conjunction with equation (1), constitutes what one may call a double Pythagorean identity. We also show that if any of the four equations above hold, then the remaining three become equivalent.

Show that the Kepler triangle satisfies equation (4).

For the Kepler triangle we have $c=\varphi,~b=1,~a=\sqrt{\varphi}$ , where $\varphi$ is the golden ratio. By definition of $\varphi$ , we have $\varphi^2=\varphi+1$ and so $c^2=a^2+b^2$ .

$\begin{equation*} \begin{split} \varphi^2&=\varphi+1\\ \implies\varphi^3&=\varphi^2+\varphi\\ (ac)^2&=\left(\sqrt{\varphi}\times\varphi\right)^2\\ &=\varphi^3\\ (ab)^2+(bc)^2&=\left(\sqrt{\varphi}\times 1\right)^2+\left(1\times\varphi\right)^2\\ &=\varphi+\varphi^2\\ \therefore (ac)^2&=(ab)^2+(bc)^2 \end{split} \end{equation*}$

Note that a random right triangle need not satisfy equation (4). For example, the very familiar case with $c=5$ , $b=4$ , $a=3$ doesn’t.

Simple characterizations

A couple of trivial equivalent statements that offer alternate descriptions of the Kepler triangle.

Suppose that a triangle satisfies equation (1). PROVE that equations (2), (3), (4) become equivalent.

In other words, the following three statements are equivalent for any right triangle (in which $c^2=a^2+b^2$ ):

the side-lengths form a geometric progression $b,a,c$
the ratio $\frac{c}{b}$ is the golden ratio
the identity $(ac)^2=(ab)^2+(bc)^2$ holds.

The result follows if we show that $(1)\implies(2)\implies(3)\implies(1)$ .

So let’s suppose that $(1)$ holds, that is, the sequence $b,a,c$ is geometric. This gives $a^2=bc$ . Since we also have that $c^2=a^2+b^2$ , it means that

$c^2=bc+b^2\implies c^2-bc-b^2=0\implies \left(\frac{c}{b}\right)^2-\frac{c}{b}-1=0$

Thus, $\frac{c}{b}$ satisfies the defining equation for the golden ratio, and so $(2)$ holds.

Next, we show that $(2)\implies(3)$ . So suppose that $\frac{c}{b}$ is the golden ratio. Then

$\left(\frac{c}{b}\right)^2-\frac{c}{b}-1=0\implies c^2=bc+b^2\implies c^2-b^2=bc$

Multiply both sides of $c^2=a^2+b^2$ by $a^2$ to get

$a^2c^2=a^2a^2+a^2b^2\implies (ac)^2=(ab)^2+(bc)^2$

since $a^2=c^2-b^2$ from the Pythagorean identity and also $c^2-b^2=bc$ from the golden ratio. Thus, $(3)$ holds.

Finally, we show that $(3)\implies(1)$ . We have $(ac)^2=(ab)^2+(bc)^2$ and $c^2=a^2+b^2$ :

$a^2(a^2+b^2)=a^2b^2+b^2c^2\implies a^4=b^2c^2\implies a^2=bc$

So the sequence $b,a,c$ is geometric, and $(1)$ is proved.

Easy equivalence.

Suppose that a triangle satisfies equation (2). PROVE that equations (1), (3), (4) become equivalent.

In other words, the following three statements are equivalent for any triangle in which $a^2=bc$ :

the triangle is a right triangle with $c^2=a^2+b^2$
the ratio $\frac{c}{b}$ is the golden ratio
the identity $(ac)^2=(ab)^2+(bc)^2$ holds.

Follow the same argument as in example 2, bearing in mind that $a^2=bc$ holds throughout in this case.

Suppose that a triangle satisfies equation (3). PROVE that equations (1), (2), (4) become equivalent.

In other words, the following three statements are equivalent for any triangle in which $\frac{c}{b}=\varphi$ :

the triangle is a right triangle with $c^2=a^2+b^2$
the side-lengths form a geometric progression $b,a,c$
the identity $(ac)^2=(ab)^2+(bc)^2$ holds.

Follow the same argument as in example 2, bearing in mind that $\frac{c}{b}=\varphi$ holds throughout in this case.

Suppose that a triangle satisfies equation (4). PROVE that equations (1), (2), (3) become equivalent.

In other words, the following three statements are equivalent for any triangle in which $(ac)^2=(ab)^2+(bc)^2$ :

the triangle is a right triangle with $c^2=a^2+b^2$
the side-lengths form a geometric progression $b,a,c$
the ratio $\frac{c}{b}$ is the golden ratio $\varphi$ .

Follow the same argument as in example 2, bearing in mind that $(ac)^2=(ab)^2+(bc)^2$ holds throughout in this case.

Similar concept

Given $\triangle ABC$ in which $\angle C=90^{\circ}$ , the Pythagorean theorem states that

(5) $\begin{equation*} AC^2+CB^2=AB^2 \end{equation*}$

(Or the usual $a^2+b^2=c^2$ .) But then the left side of equation (5) is just the sum of the squares of two altitudes — the altitude from vertex $A$ and the altitude from vertex $B$ — and so we can re-write (5) as:

(6) $\begin{equation*} h_A^2+h_B^2=AB^2 \end{equation*}$

where $h_A$ and $h_B$ are the altitudes from $A$ and $B$ . The latter equation expresses the fact that in a right triangle, the sum of the squares of two altitudes is equal to the square of the third side (the hypotenuse). We’ll use this intrinsic characteristic of right triangles as the defining condition for the triangles under consideration in the second part of today’s discussion.

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

if, and only if, $a^2+b^2=4R^2$

, where $R$

is the circumradius of triangle $ABC$

First suppose that $h_A^2+h_B^2=AB^2$ . Since $h_A=\frac{bc}{2R}$ and $h_B=\frac{ac}{2R}$ , we get

$\left(\frac{bc}{2R}\right)^2+\left(\frac{ac}{2R}\right)^2=c^2\implies a^2+b^2=4R^2$

Conversely, suppose that $a^2+b^2=4R^2$ . Then:

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

QED: $h_A^2+h_B^2=AB^2\iff a^2+b^2=4R^2.$

PROVE that $\cos C=0$

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

, if $a^2+b^2=4R^2$

Begin with $R=\frac{c}{2\sin C}$ as per the extended law of sines. The rest is:

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

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

, if a non-right triangle satisfies $a^2+b^2=4R^2$

Since the given triangle is not a right triangle, we take $\cos C=\frac{2ab}{a^2+b^2}$ from the preceding example. However, since $\cos C$ is always equal to $\frac{a^2+b^2-c^2}{2ab}$ in any triangle, we have:

$\begin{equation*} \begin{split} \frac{2ab}{a^2+b^2}&=\frac{a^2+b^2-c^2}{2ab}\\ (a^2+b^2)\Big(a^2+b^2-c^2\Big)&=4a^2b^2\\ (a^2+b^2)^2-4a^2b^2&=(a^2+b^2)c^2\\ (a^2-b^2)^2&=(ac)^2+(bc)^2 \end{split} \end{equation*}$

Similar to equation (4) for the Kepler triangle. Let’s take a shot at making it exact.

Consider a non-right triangle that satisfies $a^2+b^2=4R^2$

. PROVE that $(ab)^2=(ac)^2+(cb)^2$

if, and only if, $\frac{a}{b}$

is the golden ratio.

First suppose that $(ab)^2=(ac)^2+(cb)^2$ . Since $(a^2-b^2)^2=(ac)^2+(bc)^2$ from the preceding example we must then have $(a^2-b^2)^2=(ab)^2$ . Positive square roots: $a^2-b^2=ab$ , or $\left(\frac{a}{b}\right)^2-\frac{a}{b}-1=0$ , and so $\frac{a}{b}$ is the golden ratio. The converse is also easy. So basically, we have non-right triangles that tend to behave like the Kepler triangle.

Special case

A sample triangle that satisfies equation (6) twice.

Consider an isosceles triangle $ABC$

in which $\angle A=120^{\circ}$

, $\angle B=30^{\circ}$

, and $\angle C=30^{\circ}.$

For this triangle we have:

$\begin{equation*} \begin{split} h_A^2+h_B^2&=AB^2\\ h_A^2+h_C^2&=AC^2\\ h_B^2+h_C^2&=\frac{1}{2}BC^2\\ h_B+h_C&=BC.\\ \end{split} \end{equation*}$

Takeaway

For any non-right triangle $ABC$ with side-lengths $a,b,c (a\neq b)$ , circumradius $R$ , and altitudes $h_A,h_B,h_C$ , the five statements below are equivalent:

$h_C=R\cos C$
$\cos C=\frac{2ab}{a^2+b^2}$
$a^2+b^2=4R^2$
$h_A^2+h_B^2=AB^2$
$(a^2-b^2)^2=(ac)^2+(bc)^2.$

In addition, if a non-right triangle satisfies any of the equivalent conditions above, then each of the following three statements implies the others:

$\frac{a}{b}$ is the golden ratio
$(ab)^2=(ac)^2+(cb)^2$
$a,\sqrt[4]{5}c,b$ is a geometric sequence.

In a nutshell, we have a non-right triangle version of the Kepler triangle.

Tasks

If a non-right triangle satisfies equation (6), PROVE that the length of the altitude from vertex $C$ can be given by $h_C=R\cos C$ .
Let be the side-lengths of a right triangle, where is the hypotenuse, and let be its circumradius. PROVE that:
- $a^2+b^2=4R^2$
- $a,c,b$ (in that order) cannot be a geometric sequence.
PROVE: For any non-right triangle with side-lengths , circumradius , and altitudes , the five statements below are equivalent:
- $h_C=R\cos C$
- $\cos C=\frac{2ab}{a^2+b^2}$
- $a^2+b^2=4R^2$
- $h_A^2+h_B^2=AB^2$
- $(a^2-b^2)^2=(ac)^2+(bc)^2.$
Suppose that a non-right triangle satisfies any of the equivalent conditions in the preceding exercise. PROVE that:
- $HO^2=5R^2-c^2$
- $AH^2+BH^2+CH^2=8R^2-c^2$
- $4(AN^2+BN^2+CN^2)=7R^2+c^2$ , where $H=$ orthocenter, $O=$ circumcenter, $N=$ nine-point center
Consider with vertices at , , and . Verify that:
- the circumradius is $R=5$
- $a^2+b^2=4R^2$ .