Given two lengths $\overline{AB} = R$ and $\overline{AE} = r$ with $R > r$, how to construct a right triangle $\triangle ABC$ with a hypotenuse $\text{length} = R$ and the length of the $\text{bisector} = r$ as shown below in the figures?

Somehow I cannot get the label working for vertex $C$, the one on the orange circle, sorry.

The right triangle $\triangle ABC$ has the angles $\alpha + \beta = \pi/2$, where $\angle ACB$ is the right angle.

The bisector $\overline{AE}$ splits $\angle CAB = \alpha$ into two.

The figure above shows the case for a relatively smaller $r$ (magenta) compared with $R$ (orange), and the figure below is for a relatively large $r$.

As an analytic-geometry problem, I have been able to solve the coordinates $(x_E, y_E)$ of point $E$ (taking e.g. point $A$ as the origin), but I cannot tell from this expression if there's a sensible compass-and-straightedge construction:

$$ x_E = \frac{r}{4R} \left( r + \sqrt{ r^2 + 8R^2} \right) \qquad \text{or equivalently} \qquad \cos \frac{\alpha}2 = \frac{1}{4R} \left( r + \sqrt{ r^2 + 8R^2} \right)$$

As of now I'm trying to understand the content of some related post here on StackExchange; upon first glance it seems considerable work is needed to make their results helpful for my case.

btw, this construction is related to the L'H$\hat{o}$pital's weight (pulley) problem (in calculus).


Here's a construction, where I'll use $s$ for the given hypotenuse:

enter image description here

  • Construct right $\triangle OAB$ with $|\overline{OA}| = |\overline{OB}| = s$.

  • Construct $\overline{AC}$ perpendicular to $\overline{AB}$, with $|\overline{AC}| = r/2$.

  • Construct $D$ where $\overleftrightarrow{BD}$ meets the "far side" the the circle about $C$ through $A$.

  • Construct $E$, the midpoint of $\overline{BD}$.

  • Construct $F$, where the circle about $B$ through $E$ meets a semicircle on $\overline{OB}$.

  • Construct $G$, the reflection of $O$ across $\overline{BF}$ (which is easy, as $\angle OFB$ is necessarily a right angle).

  • Construct $H$ where $\overline{BG}$ meets the semicircle on $\overline{OB}$.

  • $\triangle OBH$ is the desired triangle.

Certainly, $\overline{BF}$ bisects angle $B$. That $|\overline{BX}| = r$ is trickier to establish: by the Power of a Point theorem, we have $$|\overline{OX}||\overline{XH}| = |\overline{BX}||\overline{XF}|$$ A bit of messy algebra shows that $$|\overline{BF}| = \frac{1}{4}\left(\; r + \sqrt{r^2 + 8s^2} \;\right) = \frac{1}{2}\left(\;\frac{r}{2} + \sqrt{\left(\;\frac{r}{2}\;\right)^2 + \left(\;s\sqrt{2}\;\right)^2\;}\;\right)$$ where the right-most expression gives the form that guided the construction. Then, the bisector has the correct length by virtue of the fact that the $\triangle OBH$ is, in fact, the solution.

I suspect there's a construction that makes all the relations clear.

  • $\begingroup$ Amazing! I had a feeling that `power of a point' might show up at some point. Thanks. There are some stuff I'm still trying to sort out, like possibly viewing this as a construct for given length from-apex-to-orthocenter $\overline{BX}=r$ in the isosceles $\triangle OBG$ of congruent sides $\overline{OB} = \overline{GB} = s$ (instead of my original proposal of given bisector in a right triangle), meanwhile I think your solution is pretty much what I'm looking for. $\endgroup$ – Lee David Chung Lin Aug 24 '16 at 5:25
  • $\begingroup$ May I ask what tool you used to make this diagram? Some features of the figure look familiar but I cannot really tell. $\endgroup$ – Lee David Chung Lin Aug 24 '16 at 5:35
  • $\begingroup$ @LeeDavidChungLin: I'm glad I could help. I use GeoGebra for my figures. $\endgroup$ – Blue Aug 24 '16 at 5:53

First, construct a right triangle with legs $r$ and $R\sqrt8$, as its hypotenuse is equal to the square root in the expression: $\sqrt{r^2+8R^2}$. Constructing $\sqrt8$ and multiplying it by $R$ are reasonably short tasks, as are the remaining two in the construction of $x_E$: adding $r$ to the hypotenuse, multiplying that by $r$ and dividing the result by $4R$. Multiplication and division can be accomplished by similar triangles, while $\sqrt8$ is the diagonal of a square of side length 2. The construction of the desired right triangle follows.

  • $\begingroup$ ah, yes, thanks, this procedure is manageable indeed. For this approach I wonder $\ldots$ how to arrange this construction relative to the final triangle to see geometrically how that length $x_E$ (as the horizontal component on $\overline{AB}$) makes $\overline{AE}$ the bisector? $\endgroup$ – Lee David Chung Lin Aug 22 '16 at 12:29
  • $\begingroup$ None that look particularly nice. It's the same thing with the construction of the heptadecagon; proving that the polygon constructed is regular requires a lot of algebra. $\endgroup$ – Parcly Taxel Aug 22 '16 at 12:39
  • $\begingroup$ ha, good point. $\endgroup$ – Lee David Chung Lin Aug 22 '16 at 12:41

Take a straight horizontal line $l$, draw a point $H$ on it and draw a segment of length $4R$ (basically intersect the line $l$ with a circle of radius $4R$ and center $H$) to the right and a segment of length $2R$ to the left. This way you get a segment $PQ \subset l$ of length $6R$ where $P$ is the left point on $l$ at distance $2R$ from $H$ and $Q$ is the right point on $l$ at distance $4R$ from $H$. Take the midpoint $M$ of $PQ$ and draw a circle $c_1$ of radius $3R$ centered at $M$ (i.e. a circle with diameter $PQ$). Now, draw the perpendicular to $PQ$ from point $H$ and let it intersect the circle $c_1$ at the point $T$ (there are two of them, just pick one, let's say the one above $PQ$). Now $HT^2 = HP\cdot HQ = 8R^2$. Next, draw on $PQ$ a point $E$ at a distance $r$ from $H$. Now, the segment $TE = \sqrt{HE^2+HT^2} = \sqrt{r^2 + 8R^2}$. Extend the segment $TE$ past $E$ and draw a point $A$ on $TE$ so that $AE = r$. Observe that $AT = AE + TE = r + \sqrt{r^2 + 8R^2}$. Now draw a circle $c_A$ centered at $A$ and of radius $4R$. Draw a line $l_T$ through point $T$ and perpendicular to line $TE$ (which is the same as line $TA$). Take one of the two intersection points of $l_T$ with $c_A$, call it $D$. By construction $$\cos{(\angle DAT)} = \frac{AT}{AD} =\frac{r + \sqrt{r^2 + 8R^2}}{4R}.$$ Draw point $B$ on the segment $AD$ so that $AB = R$. Draw the line $BE$ and construct the orthogonal projection $C$ of point $A$ on the line $BE$. Triangle $ABC$ is your triangle.

One way to construct point $C$ and to verify the construction is the correct one, is to look again at the intersection of circle $c_A$ of radius $4R$ and the line $l_T$ passing through point $T$ and perpendicular to line $AT$. One intersection point we already selected and called $D$ and let the other be $D'$. Then triangle $ADD'$ is isosceles with $AD=AD'$. Draw the altitude $DC'$ where $C'$ is on $AD'$. Define by $E'$ the intersection point of $AT$ and $DC'$ (i.e. it is the orthocenter of $ADD'$). The point $C'$ is also the second intersection point of $AD'$ and the circle circumscribed around right-angled triangle $ACT$ which is also the circle of diameter $AD$.

Now, we want to make sure that $AE' = 4r$ (so that the 4 times smaller segment $AE$ is of length $r$). By construction $AE'$ is an angle bisector of angle $\angle DAC'$. Now let for simplicity denote $$AT = h = r + \sqrt{r^2 + 8R^2} \,\, \text{ and by } \,\, AD = AD' = a = 4R.$$ Observe that $\angle DAT = \angle D'AT$ since $AT$ is the angle bisector of $\angle DAD'$. Furthermore $\angle E'TC' = \angle E'DT$ because $C'ADT$ is inscribed in a circle, or also because the right-angled triangles $TDE'$ and $C'AE'$ are similar. Thus $ \angle TAD = \angle TDE'$ which means triangles $TAD$ and $TDE'$ are similar so $$\frac{AD}{AT} = \frac{DE'}{DT}.$$ Follow the calculations \begin{align} DT &= \sqrt{a^2 - h^2} \,\, \text{ and } \,\, DE' = \sqrt{(h-AE')^2 + a^2-h^2}\\ \frac{AD}{AT}&=\frac{DE'}{DT} = \frac{a}{h} =\frac{\sqrt{(h-AE')^2 + a^2-h^2}}{\sqrt{a^2 - h^2}}. \end{align}
Square the last equation on both sides \begin{align} &\frac{a^2}{h^2} = \frac{(h-AE')^2 + a^2-h^2}{a^2 - h^2} = \frac{(h-AE')^2}{{a^2 - h^2}} + 1,\\ &(h-AE')^2 = \left(\frac{a^2}{{h^2}} - 1\right)(a^2-h^2) = \frac{(a^2-h^2)^2}{h^2}\\ &AE' = h - \frac{a^2-h^2}{h} = \frac{2h^2 - a^2 }{h} \end{align}
Plug the original expressions for $a$ and $h$ in the latter equation and you get \begin{align} AE' &= \frac{2 (r^2 + r^2 + 8R^2 + 2r\sqrt{r^2+8R^2}) - 16R^2}{r+\sqrt{r^2+8R^2}}\\ &= \frac{4r^2 + 4r\sqrt{r^2+8R^2}}{r+\sqrt{r^2+8R^2}}\\ &=4r. \end{align}

  • $\begingroup$ Thank you. I'm thinking that this arrangement suggests the construction of an isosceles (with base on $l_T$ and half of it being $\triangle DAT$) that naturally has the vertex angle bisected, instead of my previous line of thought involving segment of length $x_E$. How this meets the requirement of lengths $R$ and $r$ is still somewhat mysterious to me, but yeah this is helpful. $\endgroup$ – Lee David Chung Lin Aug 22 '16 at 18:11
  • $\begingroup$ That is clearly true. If $D'$ is the second point of intersection of circle $c_A$ and $l_T$ (the other one being $D$) then $ADD'$ is isosceles triangle with angle the angle of the right angled triangle you are after. Then the altitude from $D$ to $AD'$ intersects $AD'$ at a point $C'$ so that the triangle $ADC'$ is a right angled triangle which is similar to the original triangle $ABC$, it's just 4 times bigger. Is this what you are trying to prove? $\endgroup$ – Futurologist Aug 22 '16 at 20:04
  • $\begingroup$ Yeah, sort of. Before you mentioned it I didn't think of this approach of having $\triangle ADC'$ that is similar but 4 times the size. I guess this can be another example of the common method in plane geometry: construct an auxiliary similar shape to illuminate some facts. $\endgroup$ – Lee David Chung Lin Aug 24 '16 at 5:09

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.