2
$\begingroup$

Suppose that I have an Euclidean triangle in the plane with side lengths $a,b,c$. Denote the angle opposite $c$ by $\theta$.

I am trying to prove, that if we keep $c$ fixed, and increase $a,b$, then $\theta$ should get smaller.

That is, if $\tilde a,\tilde b,c$ are side lengths of another triangle, and $\tilde \theta$ is the corresponding angle, then

$$ \tilde a \ge a, \, \tilde b \ge b \Rightarrow \tilde \theta \le \theta.$$

I tried to prove this via the law of cosines:

$$ \cos(\theta)= \frac{a^2+b^2-c^2}{2ab}\le \frac{\tilde a^2+\tilde b^2-c^2}{2\tilde a \tilde b}=\cos(\tilde \theta),$$

but somehow got stucked.

Is there a simple algebraic proof of this inequality? or alternatively, geometric proof?

$\endgroup$
2
$\begingroup$

This is a simple counterexample:

enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks. I also realized the statement doesn't always hold. But it does hold whenever all the angles are less than $\frac{\pi}{2}$ (by the sine law). And perhaps also when $\theta$ itself is bigger than $\frac{\pi}{2}$. BTW, may I ask how did you produce this example? $\endgroup$ – Asaf Shachar Apr 22 at 15:49
  • 1
    $\begingroup$ @Asaf Shachar: this example was prepared using Asymptote, "Asymptote is a powerful descriptive vector graphics language that provides a natural coordinate-based framework for technical drawing. Labels and equations are typeset with LaTeX, for high-quality PostScript output." Also, you can check out this forum. $\endgroup$ – g.kov Apr 22 at 15:55
0
$\begingroup$

The way I understood the given problem which to me makes sense:

"If we keep positions of $(A,B)$ fixed and at least one of lengths of $(a,b)$ increased, then show that $\theta$ included between them decreases."

enter image description here

An ellipse is a convenient choice since adjacent legs sum up to a constant. Since confocal ellipses do not intersect it suffices to choose on layer as shown: $$h= a \sin \beta=b \sin \alpha\,;\tag1$$ Lawof Sines $$ \frac{\sin \theta}{c}=\frac{\sin \alpha}{a}=\frac{\sin \beta}{b}=\frac{\sin \alpha+\sin \beta}{a+b}=\frac{\sin \alpha+\sin \beta}{p}\tag2$$ where $p$ is major axis $>c$.

Since $\beta=\alpha-\theta$ $$\frac{\sin (\theta-\alpha) +\sin \beta}{\sin \theta}=\frac{p}{c}>1 \tag3$$

$\dfrac{p}{c} $ will be always greater than $1$ and expression at left increases monotonously with $\theta$, proving the proposition... the way I understood it.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.