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We are given a circle with radius $1$, its center point and an inscribed isosceles triangle with $AB=AC$ and its height (as shown in the picture below). Can we express the area $(ABC)$ as a function of $θ$ where $θ=B \hat{A} C$?

Isosceles triangle inscribed in a circle.

How I tried:

I put the diagram in a Cartesian coordinate system.

Same diagram on a coordinate grid with circle centered at the origin.

And since the circle is of radius $1$ the points are $A(0,1),\, B(\cos x_1, \sin x_1),\, C(\cos x_2, \sin x_2)$.

But then since we don't know where exactly points $B,C$ are in the circle we can't say that they have the same $y$ value and make the substitution $\sin x_1 = - \sin x_2$ or $\cos x_1 = \cos x_2$.

How can I proceed? Any help would be appreciated.

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    $\begingroup$ We can WLOG supposed that $B$ and $C$ have the same $y$-coordinates by simply choosing the $x$-axis to be parallel to $BC$. $\endgroup$
    – IraeVid
    Apr 16 at 15:43
  • $\begingroup$ Chords $AB = AC$, so their angles subtended at the centres are equal: $\angle AOB = \angle AOC$. After you WLOG chose $A=(0,1)$, then for $B$, angle $x_1 = \frac \pi2+\angle AOB$, while for $C$, angle $x_2 = \frac\pi2 - \angle AOC$. The $y$-coordinates of $B$ and $C$ are equal: $$\sin x_1 = \sin \left(\frac \pi2+\angle AOB\right) = \sin \left(\pi - \frac \pi2+\angle AOC\right) = \sin(\pi-x_2) = \sin x_2$$ $\endgroup$
    – peterwhy
    Apr 16 at 16:15

4 Answers 4

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Note that since $\triangle ABC$ is an isosceles triangle, $AD$ is the angle bisector of $\angle A$ ( It follows from $\triangle ADB \cong\triangle ADC$). Connect the center of the circle to $B$ and $C$.

enter image description here

We have $\angle OAC= \angle OCA= \dfrac{\theta}2$, Hence $\angle COD= \theta$. Applying the cosine rule for $\triangle AOC$,

$$AC^2=OC^2+OA^2-2\times OA\times OC\times \cos(\pi-\theta)$$ $$AC^2=2+2\cos(\theta)\quad\Rightarrow\quad AC=AB=\sqrt{2(1+\cos\theta)}$$

Now by the Sine Rule, the area of $\triangle ABC$ is,

$$\frac12AB\times AC\times\sin\theta=\sin\theta(1+\cos\theta)=\sin\theta+\frac12\sin(2\theta)$$

Edit:

As mentioned by @peterwhy, from $\angle COD= \theta$ one can directly calculate the area by noticing $AD= 1+\cos\theta$ and $BC= 2\sin\theta$.

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    $\begingroup$ By the way, in your formula for the area of $\triangle ABC$, $\sin\theta = BD = DC$, and $1 + \cos \theta = OA + OD$. $\endgroup$
    – peterwhy
    Apr 16 at 15:48
  • $\begingroup$ Re finding $\angle COD$: Note that this also follows directly from the inscribed angle theorem and the fact that $AD$ is an angle bisector of $\angle BOC$. I found that more straightforward to internalize, but both proof methods work. $\endgroup$
    – Kevin
    Apr 17 at 16:06
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Other answers specifically respond to your question about how to proceed with cartesian coordinates. However, imposing coordinates is not necessary, and so this answer produces formulas for the area without them, using only the definitions of the trigonometric functions and the inscribed angle theorem.

First, observe that there are three pieces of information:

  • Circle radius $r$,
  • Isosceles triangle height $h$, and
  • Angle $\theta$.

They are not independent. In fact, any two of them determines the third one via the relationship $$ \frac{h}{r} = 1 + \cos\theta. $$ As a result, there are three ways to answer the question, depending on which pair of quantities we take as given.


Height $h$ and angle $\theta$ given.

Since the triangle is isosceles, each half is a right triangle, and it's easy to see that $$ \frac{\frac12 b}{h} = \tan\bigl(\tfrac12\theta\bigr) $$ or $$ b = 2h\tan\bigl(\tfrac12\theta\bigr), $$ where $b$ is the base of the triangle perpendicular to height $h$. Then the area is $$ A = \tfrac12 bh = h^2\tan\bigl(\tfrac12\theta\bigr). $$

But, we can do better since there is a nice half-angle formula for the tangent function: $$ \tan\bigl(\tfrac12\theta\bigr) = \frac{\sin\theta}{1 + \cos\theta}. $$

Thus, $$ A = \frac{h^2\sin\theta}{1 + \cos\theta}, \tag{$h$ and $\theta$} $$ which is a formula that works without specifying the radius $r$ explicitly.


Radius $r$ and angle $\theta$ given.

Since $h$ and $\theta$ together do specify the radius $r$ by the formula $$ h = r(1 + \cos\theta), $$ the area formula can also be rewritten as $$ A = \frac{r^2 (1 + \cos\theta)^2 \sin^2\theta}{1 + \cos\theta} = r^2 \sin\theta\,(1 + \cos\theta). \tag{$r$ and $\theta$} $$


Height $h$ and radius $r$ given.

In this case, we derive the following relationship among the variables: \begin{align} r\sin\theta &= r\sqrt{1 - \cos^2\theta} \\ &= \sqrt{r^2 - r^2\cos^2\theta} \\ &= \sqrt{r^2 - (h - r)^2}, \end{align} so the area is \begin{align} A &= \frac{h^2 \, r\sin\theta}{r(1 + \cos\theta)} \\ &= \frac{h^2 \sqrt{r^2 - (h - r)^2}}{h} \\ &= h \sqrt{r^2 - (h - r)^2}, \end{align} or $$ A = h \sqrt{h(2r - h)}. \tag{$h$ and $r$} $$

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The complete law of sines is, for circumradius $R$:

$2R=\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}$

You have all the angles since it's isosceles and you control the apex angle, and your circumradius is $1$ by definition, so you have all side lengths via the law of sines:

$a=2R \sin{A}, b=2R \sin{B}, c=2R \sin{C}$

Finally, area$_{abc}=\frac{abc}{4R}$.

This result derives from area$_{abc}=\frac{1}{2}bh=\frac{1}{2}ab\sin{C} = \frac{1}{2}\frac{abc}{2R}=\frac{abc}{4R}$

Putting that all together, and supposing $x$ is the apex angle:

$f(x)=\frac{1}{4R}(abc)=\frac{(2R)^3}{4R}(\sin{x}\ \sin^2\frac{\pi-x}{2})=2R^2(\sin{x}\ \sin^2\frac{\pi-x}{2})$

Edit: to link the two answers so far, note that $\sin{\frac{\pi-x}{2}}=\sin({\frac{\pi}{2}-\frac{x}{2}})=\cos{\frac{x}{2}}$, so $2R^2(\sin{x}\ \sin^2\frac{\pi-x}{2})= 2R^2(\sin{x} \cos^2\frac{x}{2})$, and using a half-angle formula $2R^2(\sin{x} \cos^2\frac{x}{2})=R^2\sin{x}(1+\cos{x})$, showing both answers as forms of the same solution.

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Here is different solution: Let $E$ be the point diametrically opposite to $A$. Then $\triangle ABD$ and $\triangle AEB$ are right triangles which are similar: they share the angle $\angle BAD = \angle BAE = \theta/2$. Then $AB = AE \cos (\theta/2)$ and $BE = AE \sin(\theta/2)$, so by similarity we have $$ AD = \frac{AB}{AE} AB = AE \cos^2 \frac\theta2 = \frac{AE}{2} (1 + \cos \theta), $$ $$ BD = \frac{AB}{AE} BE = AE \sin \frac\theta2 \cos \frac\theta2 = \frac{AE}{2} \sin \theta. $$ Now $AE = 2$, so the answer is $$ \operatorname{Area}(\triangle ABC) = 2 \operatorname{Area}(\triangle ABD) = AD \cdot BD = \Bigl(\frac{AE}{2}\Bigr)^2 \sin \theta \, (1 + \cos \theta) = \sin \theta \, (1 + \cos \theta). $$

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