Do these triangles with parallel hypotenuses?

This originated as a question that I encountered while woodworking, but nerdy me had to try and see if there was a solution. Say I have two right triangles, and each of them have the same angles, but one is slightly larger than the other (with the right angle for both at the origin). Arranging the triangles on top of one another in this way means that the hypotenuses are parallel and separated by a certain amount. If I pick the most acute angle and name it theta, I know and can express three specific pieces of information (in addition to the obvious piece that one angle is 90 degrees for each triangle):

1. The length of the opposite side of the larger triangle is 0.5.
2. The length of the adjacent side of the smaller triangle is 1.5.
3. The separation between the two hypotenuses is 0.125.

The question is, can I solve the remaining pieces of information to fully define these triangles? I know that you generally need three pieces of info to solve for any given triangle, and I don't have that for each individual triangle, I only have two pieces of information per triangle. But I do have a fixed and known relationship between the two the separation of the parallel hypotenuses) which I feel like I ought to be able to use in some way as my third piece of information. For the life of me though, I can't figure out how to do it. Any ideas? Am I trying to solve the unsolvable? Is there a way to do this? Thanks in advance for the help!

• It would be good if you could include a diagram to show what you mean. Oct 8 '20 at 8:23

Say the larger triangle is $$AOB$$ with $$A$$ on the $$+y$$-axis, $$O$$ at the origin and $$B$$ on the $$+x$$-axis, and similarly for the smaller triangle $$A'OB'$$. We have $$AO=0.5$$ and $$OB'=1.5$$. Now let $$A'O=x$$; we then have by similar triangles $$\frac{1/2-x}{1/8}=\frac{\sqrt{x^2+9/4}}{3/2}$$ This is a quadratic equation. Solving: $$6-12x=\sqrt{x^2+9/4}$$ $$36-144x+144x^2=x^2+9/4$$ $$143x^2-144x+135/4=0$$ $$x=\frac{144-3\sqrt{159}}{286}=0.3712288\dots\qquad(x<1/2)$$ From there, since we know the triangles are similar, all other data can be computed.

• Your solution should be correct. I have verified that your solution gives around the right value of $x$ on GeoGebra. Oct 8 '20 at 8:52
• Can you share what the base formula is to generate the equation you started with? I understand all the math after that, and I validated with my woodworking project that 0.3712... is in fact the length of the small triangle short side, but I have no idea how you got that equation to start with... apologies in advance if that should have been obvious, I am in no way a studied mathematician lol... Oct 8 '20 at 9:38
• @Valiant See diagram at cdn.discordapp.com/attachments/247802651096514560/… and remember the Pythagorean theorem too! (All triangles in here are similar, actually.) Oct 8 '20 at 9:41
• That diagram is what I have on my paper here, but I don't understand why you took the data from that diagram and arranged it in the equation you made the way you did. I know it works, because I measured it, but I don't know how you got there. From what I deduced so far, the left side is the difference between the vertical sides of the two triangles divided by the 0.125 separation, and the right side is small triangle hypotenuse divided by the horizontal side of the small triangle. I have no idea why you put those values there, is what I am asking, apologies if I'm not explaining properly... Oct 8 '20 at 9:54
• @Valiant No! The numerator on each side represents the ratio of hypotenuse to adjacent, which should be the same since the triangles are similar. The LHS is the ratio for the tiny triangle near $A$, and the RHS is the ratio for $A'OB'$. In particular, the separation of 0.125 is the adjacent of the tiny triangle. Oct 8 '20 at 9:56

In the general case:

$$b\cos\theta = d + a\sin\theta \tag{1}$$

Knowing any three of $$a$$, $$b$$, $$d$$, $$\theta$$, you can find the fourth. Of course, $$\theta$$ is the tricky one, but we can write $$(d+a\sin\theta)^2=b^2\cos^2\theta\quad\to\quad d^2+2a d\sin\theta+a^2\sin^2\theta=b^2(1-\sin^2\theta) \tag{2}$$ Solving the quadratic in $$\sin\theta$$, we get

$$\sin\theta = \frac{-ad\pm b\sqrt{a^2+b^2-d^2}}{a^2+b^2} \tag{3}$$

For non-negative acute $$\theta$$ (and non-negative $$a$$ and $$b$$), we take the "$$\pm$$" to be "$$+$$".

As a woodworker, you would probably prefer to know $$\tan\theta$$. A little work yields

$$\tan\theta = \frac{ab-d\sqrt{a^2+b^2-d^2}}{(a+d)(a-d)} \tag{4}$$

For the specific case described in the question, we have $$a=3/2$$, $$b=1/2$$, $$d=1/8$$, so $$\tan\theta = \frac{1}{143} (48 - \sqrt{159}) = 0.2475\ldots \quad\to\quad \theta\approx 13.9^\circ \tag{5}$$

Let $$\angle BEA$$ be $$\theta$$. Then $$\angle FDE = \theta$$ by similarity, so $$\sin \theta = \frac{0.125}{ED} \Rightarrow ED = \frac{0.125}{\sin \theta}$$.

Similarly, $$\angle BCG = 90º - \theta$$, so $$\angle CBG = \theta$$ as well, and $$\cos \theta = \frac{0.125}{BC} \Rightarrow BC = \frac{0.125}{\cos \theta}$$.

Since $$\Delta ABE \sim \Delta ACD$$, we have that:

$$\frac{AB}{AE} = \frac{AC}{AD} \Rightarrow \frac{0.5 - 0.125/\cos \theta}{1.5} = \frac{0.5}{1.5 + 0.125/\sin \theta}$$ $$\Rightarrow \sin \theta \cos \theta (0.5 - 0.125/\cos \theta)(1.5 + 0.125/\sin \theta) = 1.5 \cdot 0.5 \sin \theta \cos \theta$$ $$\Rightarrow \sin \theta \cos \theta (0.75 + 0.0625/\sin \theta - 0.1875/\cos \theta - 0.015625/ (\sin \theta \cos \theta)) = 0.375 \sin \theta \cos \theta$$ $$\Rightarrow 0. 75 \sin \theta \cos \theta + 0.0625 \cos \theta - 0.1875 \sin \theta - 0.015625 = 0.375 \sin \theta \cos \theta$$ $$\Rightarrow \theta \approx 13.9º$$

which gives $$AB, AC, AE, AD$$, and you can find the hypotenuses $$BE$$ and $$CD$$ using Pythagoras.