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An isosceles triangle $ABC$ with duplicated sides $AC=BC$ of length $2$ each (so $AC=BC=2$) and the third side $AB$ has length $1+\sqrt{5}$. I am looking for the angle $\angle ACB$ without using trigonometry.

I know that the ratio $AC/AB$ is the reciprocal of the golden ratio, so this triangle is very much related to the golden triangle.

My plan to solve this problem is by bisecting angle $\angle CAB$ and to get two similar triangles (a similar approach used in golden triangle).

Let $D$ be the point on $CB$ where $AD$ is the angle bisector of $\angle CAB$. I just need to show that $\angle ADB=\angle ABD$, then I could finish up the rest by showing that triangle $\Delta ABC$ is similar to triangle $\Delta BDA$.

Showing $\angle ADB=\angle ABD$ is the point where I got stuck. Any clue on how to show that?

I am also curious whether there is any other way to solve this problem without using trigonometry and golden ratio?

Many many thanks.

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One way to find $\angle{ACB}$ using similar triangles :

Let $E$ be a point on the extension of the line $AC$ such that $\angle{CEB}=\angle{CAB}$.

Since $\triangle{AEB}$ is similar to $\triangle{ABC}$, we have $$AE:AB=AB:AC\implies AE=3+\sqrt 5$$ from which $CE=AE-AC=1+\sqrt 5$ follows.

It follows that $\triangle{CBE}$ is an isosceles triangle with $EC=EB$.

Let $\alpha=\angle{CAB}$. Then, we have $\angle{CAB}=\angle{CBA}=\alpha$ and $\angle{ACB}=\angle{CEB}+\angle{EBC}=3\alpha$.

Therefore, $180^\circ=5\alpha\implies \alpha=36^\circ\implies \angle{ACB}=3\alpha=108^\circ$.

$\qquad\qquad\qquad $enter image description here


Showing $\angle ADB=\angle ABD$ is the point where I got stuck. Any clue on how to show that?

We have $\angle{ADB}\not=\angle{ABD}$.

Proof :
Suppose that $\angle{ADB}=\angle{ABD}=2\theta$. Let $\theta=\angle{DAB}=\angle{CAD}$. Then, considering $\triangle{DAB}$, we see that $\angle{CDA}=\angle{DAB}+\angle{DBA}=3\theta$. Also, considering $\triangle{CAD}$, we see that $\angle{ACD}=\angle{BDA}-\angle{CAD}=\theta$. Now, considering $\triangle{ABC}$, we have $180^\circ=5\theta\implies \theta=36^\circ$. Having $\angle{CAB}\gt\angle{ACB}$, we have $BC\gt AB$, i.e. $2\gt 1+\sqrt 5$ which is impossible. $\quad\blacksquare$


Another way to find $\angle{ACB}$.

The isosceles triangle can be found in a regular pentagon with side length $2$.

Let $x$ be the length of a diagonal of the pentagon $PQRST$.

Then, by Ptolemy's theorem, we have $$|\overline{QR}|\times |\overline{PS}|+|\overline{PQ}|\times |\overline{RS}|=|\overline{PR}|\times |\overline{QS}|$$ $$\implies 2x+4=x^2\implies x=1+\sqrt 5$$

It follows that $\angle{ACB}=108^\circ$.

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Given

$$ \sin \angle ACM = \dfrac{(\sqrt5 +1)/2}{2} $$

This is recognized as half angle of regular pentagon associated with GoldenRatio $\varphi$

$$ \angle ACM= 54^{\circ}$$

Double this angle

$$ \angle ACB= 108^{\circ}.$$

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