Construction of a isoceles Triangle

Question

How can one construct an isosceles triangle with ruler and compass with the following givens

1. the sum of the base and a side
2. the head angle

Answer 1

Split your isosceles triangle in half. You now have a right angled triangle ($c$ is the hypotenuse). Call the head angle $\theta$ The head angle has been halved, so replace $\theta$ with $\frac{\theta}{2}$. Then, subtract $\theta$ from 90 to get the angle opposite the head angle, but not the right angle. Call the other sides $b$ and $a$, with $b$ being the base. $b + c = x$ . $c = x-b$ _______ $a^2 + b^2 = (x-b)^2$ _______________ $a^2 + b^2 = x^2 +b^2 - 2bx, a^2 = x^2 - 2bx$.

Sine rule : $$\frac{b}{\sin {\frac{\theta}{2}}}=\frac{x-b}{\sin (90-{\frac{\theta}{2})}}$$ because $c = x-b$ $$\frac{x-b}{\sin (90-{\frac{\theta}{2})}}=\frac{x-b}{\cos \frac{\theta}{2}}$$

Because of the trig identity $sin(90-x)\equiv cos(x)$

$$\frac{x-b}{\cos \frac{\theta}{2}}=\frac{b}{\sin {\frac{\theta}{2}}}$$

By multiplying by $\cos(\theta/2)$ and dividing by $b$,

$$\frac{\cos\frac{\theta}{2}}{\sin{\frac{\theta}{2}}} = \frac{x-b}{b} = \cot{\frac{\theta}{2}} $$

because of the identity $\frac{\cos x}{\sin x} \equiv \cot x$

$$\frac{x-b}{b} = \cot{\frac{\theta}{2}} $$

If we expand the LHS, we get $\frac{x}{b} +\frac{-b}{b} = \frac{x}{b} - 1$

Now, if we add $1$, we get:

$$\frac{x}{b} = \cot (\frac{\theta}{2}) +1$$

Take the reciprocal, and we get:

$$\frac{b}{x} = \frac{1}{\cot(\frac{\theta}{2})+1}$$

And finally, multiply be $x$

$$b = \frac{x}{\cot(\frac{\theta}{2})+1}$$

With this, you can find other side lengths, to make it suitable to construct!

Answer 2

As more than one comment has pointed out, your triangle is fully determined by your givens.

Let your base be $b$ and the other equal sides each be $a$. The vertex ("head") angle is $\theta$.

The base angles are each $\displaystyle \frac{\pi}2 - \frac 12 \theta$.

By the Sine Rule,

$$\frac a{\sin(\frac{\pi}{2} - \frac 12 \theta)} = \frac b{\sin\theta}$$

$$2a\sin\frac 12 \theta \cos \frac 12 \theta = b\cos \frac 12 \theta$$

The cosine terms can be cancelled if we disregard the degenerate case of $\theta = \pi$, so

$$b = 2a \sin\frac 12 \theta$$

Now you're given the sum of $a$ and $b$, let's call that $s$. So $b = s - a$.

Solving for $a$, we get:

$$a = \frac s{1 + 2\sin\frac 12 \theta}$$

which will immediately allow you to construct the required triangle with a ruler and protractor.

Answer 3

Given the length of AP (= AY + YP as a whole) and $\angle A$. The object is to locate X (and therefore Y) such that $\triangle AXY$ is isosceles.

1. Construct AA’ = the angle bisector of $\angle A$.

2. Form the rhombus APA’P’.

3. Let the angle bisector of $\angle APP’$ cut AP’ at X. Then, $\angle 1 = \angle 2$

Proof: XY should be a line parallel to PP’. Then, $\angle 1 = \angle 3$.

$\angle 3 = \angle 2$ implies XY = YP.

Answer 4

Let's call the angle $\theta$

Let's call the sides length $a$ and the base length $b$. For convenience let's say that $a+b=1$ unit. (Doesn't matter what unit so we just define $1 = a+ b$)

With $\theta$ we can construct arbitrary similar isosceles triangles of arbitrary size. We can construct one where the side is $1$. This triangle has a scale of $1/a$ compared to the triangle we want. ($1 = a*\frac 1a$). So the base is $b*\frac 1a = \frac ba$ in length.

We can construct the base angles, so we can construct a base of length $1$ and the two base angles to construct a similar isosceles triangle with base $=1$. This triangle will have a scale of $1/b$. ($1 = b*\frac 1b$). So the sides of this triangle is $a*\frac 1b = \frac ab$ in length.

So we can construct lengths of $1$, $a/b$ and $b/c$ and we need to figure out how to construct a length of $a$ or $b$ from these.

Now $a = \frac a1 = \frac a{a+b} = \frac 1{1 + b/a}$.

Constructing a segment of length $1 + b/a$ is simple extension.

If we have a length $m$ and length $1$ we can construct length $1/m$ like this:

Make a segment of $\overline {AB}$ of length $m$. Make a perpendicular line at $B$. Mark off point $C$ at a distance of $1$. So $ABC$ is a right triangle. $AB = m$ and $BC=1$. Construct an perpendicular to $\overline {BC}$ at $C$. Extend it until it intersects $\vec{AB}$ at point $D$.

Triangles $ABC$ and $CBD$ are similar. And $\frac {AB}{BC} = \frac {BC}{BD}$ and so $BD = BC*\frac {BC}{AB} = 1*\frac 1m = \frac 1m$.

So lets put this all together:

1) Construct $\angle \theta$ at point $O$ and mark off $A$ and $B$ so that $OA = OB= a+b$ and $\angle AOB = \theta$. This is an isosceles triangle proportional to the one we want.

$AB = m = \frac {b(a+b)}a$.

1) Construct $\angle \theta$ at point $O$ and mark off $A$ and $B$ at any distance so that $OA = OB$ and $\angle AOB = \theta$. This is an isosceles triangle proportional to the one we want.

2) Elsewhere construct a line $\overline {CD}$ of length $m = AB$. Extend it to $CE$ so that $CD = CD + a+b = m + a+ b= (a+b)(1 + \frac ba)$.

3) Elsewhere construct a right triangle $FGH$ so that $FG = CE = (a+b)(1 + \frac ba)$ and $GH= a+b$.

4) Construct triangle $HGI$ so that $\angle FHI$ is right, and $I$ is the point of intersection with $\vec {FG}$.

$\triangle FGH$ is similar to $\triangle HGI$ and $\frac {GI}{HG}= \frac {HG}{FG}$ and $GI = HG^2\frac 1{FG} = (a+b)^2\frac {1}{(a+b)(1 + \frac ba)}=\frac{a+b}{1 + \frac ba}= a$

5) Elsewhere construct triangle $JKL$ so that $JK = GI = a$ and $\angle JKL = \theta$ and $KL = GI = a$.