The distance between David and Eric

Question

Alice (A) and Bob (B) are facing each other. They both turn $10$ degrees and now they can directly see Claire (C). If Alice turns another $10$ degrees in the same direction, she will see Eric (E). If Bob turns another $20$ degrees in the same direction, he will see David (D). But Alice can't see David because Claire is in the way. Bob can't see Eric because David is in the way.

The distance from Claire to Eric is $5$ meters. What is the distance from David to Eric?

The figure above and where people are shown might not be totally right.

Note: This problem doesn't require a computer for its solution; it can be solved manually. However, you're free to use a computer to sketch out the diagram.

Answer 1

No calculator is required as OP suggested:

Firstly, we need to draw the story:

Alice and Bob are looking at each other, both turn $10^o$ and now they both can directly see Claire.

So the only way they look at the same person, if one of them turns counterclockwise and the other clockwise since they were originally looking at each other as shown above and the intersection has to be our girl Claire.

If they continue turning in their same directions before, Alice will directly able to see Eric after turning $10^o$ more, and Bob will directly able to see David after turning $20^o$ more. But Alice cannot see David because Claire blocks the view, and Bob cannot see Eric because David blocks the view.

So they continue turning, as it is shown above, since A cannot see D because of C, only possible place to put D is inbetween E and D because B cannot see E because of D. Everything seems okay. So the question becomes a hard (in my opinion) geometry question. It requires somehow finding the angles around $\triangle ECD$ and the given data is not enough yet. But we know something from the graph that is $\left | AC \right |=\left | BC \right |$. After working on it for some time, I have found something interesting which requires another drawing as you see below:

So If have draw another line starting from C which is equidistance to $\left | AC \right |$ and $\left | BC \right |$. We can create two more isosceles triangles to play with. and I decided to draw it as you see above and call the intersection point on $\vec{BDE}$ as $F$, or Felicia. In other words, I draw a line with the same angle

$\measuredangle CBE=\measuredangle CFE=20^o$

So there are $3$ isosceles triangles. So let's see what we can find from these triangles:

It is known that $\measuredangle AEF=50^o$ since A turns $10^o$ twice and B turn $10^o$ and $20^o$ degrees. the "x"s are coming from $\left | FC \right |=\left | AC \right |$ so if we check the triangle $\triangle AFE$ and sum the angles we got:

$(\measuredangle FAE=x-10)+(\measuredangle AFE=x+20)+(\measuredangle FEA=50)=180$

from here x can be found as

$60^o$ so $\measuredangle FAE=50^o$

so $\measuredangle FAE=\measuredangle AEF$ and $\left | AF \right |=\left | FE \right |$. Moreover, $\measuredangle AFC$ angle becomes $60^o$ degrees with simple triangle geometry, likewise $\measuredangle CAF$ where I call red angles with $x$s. As a result of this draw, we got

A equilateral triangle, $\triangle CAF$ with lots of equal lines as you see below:

After checking all angles, It can be easily found that all red lines become equal to each other and we got

Another isosceles triangle: $\triangle CEF$

To be honest, the rest is just filling the blank angles but if we go further by filling the angles $\measuredangle FEC$ and $AEC$ which are critical to find the angles around the triangle $\triangle CED$:

$\left | CE \right |=\left | ED \right |$ since the angle $\measuredangle ECD=\measuredangle EDC=40^o$, in other words:

The distance between David and Eric is $5$ meters also, no calculator is used, just used a software for drawing.

Answer 2

The answer is:

$5$ meters.

---

The actual to-scale figure is something like this:

Here $\angle CAB=\angle EAC=\angle CBA=10^\circ$ and $\angle EBC=20^\circ$. Now looking at this computer-generated, accurate diagram suggests that

$EC=ED$; we'll prove precisely that. It's not hard to see that condition translates into $\angle CED=100^\circ$ (a proof will be given later on). For now, assume $\angle CED=x$, and since $$\angle AED=180^\circ-20^\circ-30^\circ=130^\circ,$$ we have $\angle AEC=130^\circ-x$.
Now using sine rule in $\triangle ECB$, $\tfrac{EC}{CB}=\tfrac{\sin 20^\circ}{\sin x}$, and similarly from $\triangle ECA$, we have $\tfrac{EC}{CA}=\tfrac{\sin10^\circ}{\sin(130^\circ-x)}=\tfrac{\sin10^\circ}{\sin(50^\circ+x)}$, and since $CA=CB$ ($\triangle CAB$ is isosceles), we have $$\frac{\sin 20^\circ}{\sin x}=\frac{\sin10^\circ}{\sin(50^\circ+x)}\iff \frac{\sin(50^\circ +x)}{\sin x}=\frac{\sin10^\circ}{\sin20^\circ}.$$Now the function $\tfrac{\sin(50^\circ+x)}{\sin x}=\tfrac{\sin x\cos50^\circ+\cos x\sin 50^\circ}{\sin x}=\cos 50^\circ+\cot x\sin 50^\circ$ is clearly decreasing on the range of values we care about, so there can be at most one value of $x$ that works. But $x=100^\circ$ does fit, because $$\frac{\sin 150^\circ}{\sin 100^\circ}=\frac{\sin10^\circ}{\sin20^\circ}\iff \sin 30^\circ\sin 20^\circ=\sin10^\circ\sin80^\circ\iff\frac12\cdot 2\sin 10^\circ\cos10^\circ=\sin 10^\circ\cos 10^\circ$$which is true.

Therefore $\angle CED=100^\circ$. Also, $\angle EDC=\angle DBC+\angle DCB=20^\circ+\angle CAB+\angle CBA=40^\circ$, so $\angle ECD=180^\circ-100^\circ-40^\circ=40^\circ=\angle EDC$ , which implies $ED=EC=5 \rm{ m}.$ $\blacksquare$

Answer 3

It turns out that:

I don't know yet.

In attempting to solve it, my conclusion was flawed. I'll leave the first steps here to help anyone else along, because maybe there's something useful in there.

WARNING: I was having trouble spoiler-tagging the explanations with the images, so nothing below this point is spoiler-tagged, if that matters to fellow solvers!

Let's start with a not-to-scale sketch, filled with what we know:

Completing sets of 180° gives us a little more information:

Now we have a few more angles we can solve for. Let's call them G, H, I, and J to simplify the next step of math.

Based on what we know of the other triangles, we have four variables and 4 equations:
G + I = 170
G + H = 180
H + J = 140
I + J = 130

We can solve these as:
G = 170 - I
H = 10 + I
J = 130 - I

Which has infinitely many solutions that could make sense in this figure based strictly on a sketch. It may be possible to determine with some geometry beyond what I'm using, but I think that without being able to calculate the sides with proper "sin(x), cos(x)" calculations, it may not be possible.

My prior solution, which is no longer conclusive:

One solution is:

G = 140
H = 40
I = 30
J = 100

If this were the solution, we could be done. Looking at Triangle CDE, we see ∠ECD = ∠EDC. In an isosceles triangle, the sides opposite each of the equal angles are equal in length. Therefore we know sides CE = DE. And since CE is 5m, we know DE must also be 5m, QED.

The problem is that {G, H, I, J} = {145, 35, 25, 105} is also a valid solution set, and it completely ruins the isosceles proof while still "looking about right." Pinning it down to the previous ideal solution would take another geometric step that I don't know right now.

Feel free to supply it if you can!

As one other element that could help people, you can use the isosceles triangle property I mentioned above to find two pairs of equal lengths in the figure:

I wasn't able to turn that into anything useful, though...

Answer 4

There are 4 equations and 4 unknowns

X + v = 170
X + y = 180
V + w = 130
Y + w = 140

Problem they are not unique. There is two sets of.

Y – v = 10

This is not to scale

Not seeing how to solve it

Any ideas?

Answer 5

(I will add figures later)

We know that

\begin{align} A\widehat{E}C + C\widehat{E}D = {130}^\circ\\ D\widehat{C}E+E\widehat{C}A = {180}^\circ\\ C\widehat{E}D+E\widehat{D}C={140}^\circ\\ E\widehat{C}A+A\widehat{E}C = {170}^\circ \end{align} Solution of this system gives us the following parametric system \begin{align} E\widehat{C}A&={180}^\circ-t_1\\ A\widehat{E}C&=t_1-{10}^\circ\\ C\widehat{E}D&={140}^\circ-t_1\\ D\widehat{C}E&=t_1 \end{align}

$t_1$ can be any angle in this interval $] 10^\circ, 140^\circ [$.

Using the law of sines we get \begin{align} \quad\frac{DE}{\sin{D\widehat{C}E}}&=\frac{CE}{\sin{E\widehat{D}C}}\\ \Leftrightarrow \quad DE &= \frac{CE}{\sin{E\widehat{D}C}}\times \sin{D\widehat{C}E} \end{align}

And we know that $CE={5}$ meters, $E\widehat{D}C={40}^\circ$ and $D\widehat{C}E=t_1$

So

\begin{equation} \boxed{DE=\frac{5}{\sin{{40}^\circ}}\times \sin{t_1}=\frac{5}{\sin{{40}^\circ}}\times \sin{D\widehat{C}E}} \end{equation}

Also $AC$ and $CB$ depends of $t_1$.