An isosceles triangle is inscribed in the unit circle. Determine the largest area of ​​the triangle can assume.

My solution attempt:

An isosceles triangle has two sides of equal size, and the sides will be at some point on the circumference of the unit circle during suggestion the $x$-axis. The height will be $1 + h \le 2$ and width $b + 1 \le 2$

Another way would be to make the vectors of the three points that will be on the periphery (assuming that it can provide the largest area on the periphery)

I really do not know how to solve it.

  • 1
    $\begingroup$ Try to show that the condition $\Rightarrow $ the triangle is equilateral. $\endgroup$ – Rohan Jan 9 '17 at 14:10


The triangle is $\Delta ABC$: $AB=AC=b$ and $BC=a$.

Also $\angle A=2x \rightarrow \angle B=\frac{\pi - 2x}{2}=\frac{\pi}{2}-x$, so:

$$S(ABC)=\frac{b^2\sin 2x}{2}$$

And by sine rule:

$$\frac{b}{\sin \angle B}=2\cdot 1 \rightarrow b=2\cdot \cos x$$

and then:

$$S(ABC)=2\cdot \cos^2x\cdot\sin 2x=(1+\cos 2x)\cdot \sin 2x$$

Can you finish?


equilateral triangle has maximum area

side would be $\sqrt{3}$ by using trigonometry

area is $\frac{\sqrt 3}{4}a^2$$=$$\frac{3\sqrt 3}{4}$

  • 1
    $\begingroup$ Without arguments this isn't worth anything. $\endgroup$ – Henrik - stop hurting Monica Jan 9 '17 at 14:57
  • $\begingroup$ for what ?clerify what you are asking. $\endgroup$ – Blaise Thunderstorm Jan 9 '17 at 15:01
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    $\begingroup$ You are making an assertion of the answer rather than determining it. To be useful, you would show how you know that the equilateral triangle is the maximum area. $\endgroup$ – Joffan Jan 9 '17 at 16:38

I assumed a central angle within the circle A to create an additional isosceles triangle and find the base N, as well as find the total height.

Because this is a unit circle, the radius is 1.

With the central angle A, we create the isosceles triangle with base N and with base angles X and with height K.

Drawing a perpendicular bisector down from the vertex of A creates two right triangles; each has height K, base angle X, upper angle (for lack of a better term) a/2, and hypotenuse of 1 (because the radius is 1). With this, we can create some functions/equations based on the variables:

X = 90 - (A/2)

sin(x) = K

cos(x) = n/2

2cos(x) = n

The total height of the isosceles triangle in question (the one with the largest area) is given by K+1.

The base of the isosceles triangle with base angle X is given by N (as is the largest possible isosceles triangle).

The area of a triangle is given by ((base*height)/2).

Therefore, we can perform substitutions, and get that

(2cos(x)(sin(x) + 1))/2 = Area.

More simply, cos(x)(sin(x) + 1) = Area.

However, we established these relations based on an assumed central angle; therefore, we need to substitute.

cos(90 - (a/2))(sin(90 - (a/2)) + 1) = Area.

Proceed from there.


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