MHB Can you solve this right triangle puzzle with a unique angle ratio?

AI Thread Summary
The discussion centers around solving a right triangle puzzle involving angles and side lengths. The triangle ABC has angle BAC defined as 2u degrees, and point D on the hypotenuse AB creates a secondary angle BCD of 3u degrees. One participant concludes that the triangle must be a 30-60-90 triangle to satisfy the conditions, leading to the calculation of u as 15 degrees. The conversation also touches on using coordinate geometry and trigonometric identities to derive relationships between the angles and side lengths. Ultimately, the unique angle ratio and the specific triangle configuration are highlighted as key aspects of the problem.
Wilmer
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B
    b-a
            D
a
                   c-b+a

C                b                  A
Right triangle ABC, with the standard a, b, c side lengths.
Angle BAC = 2u degrees.

Point D is on the hypotenuse AB, such that:
BD = b-a, angle BCD = 3u degrees.

Calculate u.
 
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Is this a problem with which you are asking for help, or is it a problem that you are posting as a challenge?
 
MarkFL said:
Is this a problem with which you are asking for help, or is it a problem that you are posting as a challenge?
Hey Mark!
It's a problem I found.
My solution to it is that it HAS to be a 30-60-90 triangle,
else the "b-a" length is impossible (combined with 2u and 3u)..
Posted it to sort of "see" if anyone agrees.

Anyhoo...if I've done something sinful, I apologize :)
 
Wilmer said:
Hey Mark!
It's a problem I found.
My solution to it is that it HAS to be a 30-60-90 triangle,
else the "b-a" length is impossible (combined with 2u and 3u)..
Posted it to sort of "see" if anyone agrees.

Anyhoo...if I've done something sinful, I apologize :)

Since you aren't sure if you are correct and are wanting to see if people agree with your solution, I would say you posted in the correct place. (Yes)
 
Let's use some coordinate geometry here and WLOG, let $a=1$...let the hypotenuse lie along the line:

$$y=1-\frac{x}{b}$$

And line segment $\overline{CD}$ lie along the line:

$$y=\cot(3u)x$$

And so we find $D$ is at:

$$(x,y)=\left(\frac{b}{b\cot(3u)+1},\frac{b\cot(3u)}{b\cot(3u)+1}\right)$$

Now, we require:

$$(b-1)^2=\left(\frac{b}{b\cot(3u)+1}\right)^2+\left(\frac{1}{b\cot(3u)+1}\right)^2$$

From this, we determine:

$$\cot(3u)=\frac{\sqrt{b^2+1}-(b-1)}{b(b-1)}$$

And using a triple-angle identity for cotangent, we have:

$$\frac{3\cot(u)-\cot^3(u)}{1-3\cot^2(u)}=\frac{\sqrt{b^2+1}-(b-1)}{b(b-1)}$$

We also know:

$$\cot(2u)=b$$

and using a double-angle identity for cotangent, we have:

$$\frac{\cot^2(u)-1}{2\cot(u)}=b$$

Form this, we determine:

$$\cot(u)=b+\sqrt{b^2+1}$$

And so we have:

$$\frac{\left(b+\sqrt{b^2+1}\right)\left(b^2+b\sqrt{b^2+1}-1\right)}{3b^2+3b\sqrt{b^2+1}+1}=\frac{\sqrt{b^2+1}-(b-1)}{b(b-1)}$$

Taking the positive real root, we find:

$$b=\sqrt{3}$$

And this in turn implies:

$$\cot(3u)=1$$

or:

$$3u=45^{\circ}$$

$$u=15^{\circ}$$
 
Thanks Mark. I'm lazy, so I went at it this way:

Assume ABC is a 30-60-90 triangle.

Let a = 1: then c = 2 and b = sqrt(3)

Since angleBAC = 2u and angleBCD = 3u,
then angleBCD = 45 and angleBDC = 75

BD = SIN(45) / SIN(75) = sqrt(3) - 1
also:
BD = a - b = sqrt(3) - 1

That's it...u=15 ! Only case where above works...

Anything wrong with that?
I'd sure go that way on a multiple choice timed test!
 
Interestingly(?) enough, seems to be only one pytagoreaner
that has these angles in a mu:nu ratio (m and n = integers).

Happens to be the 3-4-5er (and its multiples, of course).
BD : DA = 1 : 4

AngleBAC = 2u and angleBCD = u

Checked 'em all where a < 10000.

File 13 stuff!
 
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