4
$\begingroup$

Let ABC be a right triangle with B the right angle. X,Y and Z are on BC, CA and AB respectively such that BXYZ is a square. If the square is of side length m, AY = r and YC = s, find m in terms of r and s.

I have two solutions for my students (I instruct a math team class): One involves proportions and the other involves the angle bisector theorem (BY is an angle bisector). I am just curious of there are any other creative solutions for this problem.

  • 0
    My students and I have done much of what's covered in math team classes: Ptolemy's Theorem, Stewarts Theorem (including a proof from Ptolemy's), Angle Bisector Theorem I and II (the latter from Stewart's), etc. I just like to show multiple solutions to my students to show them how to think creatively. I was just curious if there are any creative solutions since mine seem brute force-ish. I am currently trying to find other solutions through Stewart's Theorem, circumscribing the triangle and extending the angle bisector to use the chord theorem, amongst other things.2012-07-23

4 Answers 4

4

Let $\theta=\angle ZYA=\angle XCY$. Then $\cos\theta=\frac{m}{r}$ and $\sin\theta=\frac{m}{s}$. It follows that $\left(\frac{m}{r}\right)^2+\left(\frac{m}{s}\right)^2=1$.

  • 0
    @Mr.Pi: I figured if it wasn't the one you meant it might be useful, since it connects with "another subject."2012-07-23
4

You described this in the comments, so perhaps you have it already, but circumscribe the circle through $\rm ABC$ with center at $\rm O$ (on $\rm AC$ since $\angle\rm B$ is a right angle) and extend $\rm BY$ to meet the circle again at $\rm M$.

Circumscribed circle

Then $\angle \mathrm{AOM}=2\angle\mathrm{ABM}$ is a right angle. Wlog I assume $\rm AB > BC$ and $s>r$. So $\mathrm{OM}=\mathrm{AO}=(s+r)/2$, $\mathrm{OY}=(s-r)/2$. Then by the Pythagorean theorem $|\mathrm{YM}|^2=(s^2+r^2)/2$, and by the intersecting chords theorem $ \mathrm{BY}\cdot\mathrm{YM} = \mathrm{CY}\cdot\mathrm{YA} \\ \sqrt{2}m\cdot \sqrt{(s^2+r^2)/2} = r\cdot s \\ m = \frac{rs}{\sqrt{s^2+r^2}} $

1

The solution for someone who really prefers algebra to geometry:

Let $AZ=p$, $XC=q$. Then the Pythagorean theorem tells us that \begin{eqnarray} p^2+m^2&=&r^2\\ q^2+m^2&=&s^2\\ (p+m)^2+(q+m)^2&=&(r+s)^2 \, . \end{eqnarray} Subtracting the first two equations from the third and simplifying yields $p+q=\frac{rs}{m}$, while subtracting the second equation from the first yields $p^2-q^2=r^2-s^2$. So $p-q=\frac{m}{rs}(r^2-s^2)$, and thus $p=\frac{1}{2}\left(\frac{mr}{s}+\frac{rs}{m}-\frac{ms}{r}\right) \, .$ Substituting this value into the first Pythagorean relation, we have $\frac{1}{4}\left(\frac{m^2r^2}{s^2}+\frac{r^2s^2}{m^2}+\frac{m^2s^2}{r^2}+2r^2-2s^2-2m^2\right)+m^2-r^2=0 \\ \frac{1}{4}\left(\frac{m^2r^2}{s^2}+\frac{r^2s^2}{m^2}+\frac{m^2s^2}{r^2}-2r^2-2s^2+2m^2\right)=0 \\ \left(\frac{mr}{s}-\frac{rs}{m}+\frac{ms}{r}\right)^2=0 \, ;$ isolating $m$ gives $m^2=\frac{rs}{\frac{r}{s}+\frac{s}{r}} \, .$

1

Rotate the triangle $XYC$ by $90$ degrees about $Y$ in such a way as to make $XY$ coincide with $YZ$. This gives a new right triangle; its legs have length $r$ and $s$, and the altitude to its hypotenuse has length $m$. So the area of this triangle is $\frac{1}{2}rs=\frac{1}{2}m\sqrt{r^2+s^2}$.