# The distance between David and Eric

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.

• This is so confusing. Sep 20, 2017 at 11:00
• Could you just show a diagram? Sep 20, 2017 at 11:46
• The answer still requires a calculator to determine, which would seem to contradict the no-computers tag. Sep 20, 2017 at 11:50
• @Seyed how did you arrive at that? Sep 20, 2017 at 20:56
• I suspected that may end up being the case, I've been trying to show that geometrically myself for hours :\ Sep 20, 2017 at 22:27

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.

• I cannot add more than two pictures because of rep...
– sila
Sep 21, 2017 at 12:43
• I got lost where you assert that △CEF is isosceles — how do you know this? (I assume because |CF|=|EF| but how you determined that is not clear to me)
– Rubio
Sep 21, 2017 at 13:55
• @Rubio FE = AF = AC = CB = CF. Each equality is true because the two sides involved form an isosceles triangle. Sep 21, 2017 at 14:01
• @Rubio I edited that part a bit, sorry for jumping that conclusion so fast.
– sila
Sep 21, 2017 at 14:14
• Great solution, welcome to PSE! Sep 21, 2017 at 15:16

$5$ 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 $ED=EC=5 \rm{ m}.$ $\blacksquare$

• how does canceling 2's imply $\sin30\sin20=\sin10\sin80$?
– JMP
Sep 21, 2017 at 11:35
• what's $\angle DBE$?
– JMP
Sep 21, 2017 at 11:46
• @JonMarkPerry $\sin 30=1/2, \sin 20=2\sin 10\cos 10$, and $\cos 10=\sin 80$. As for $DBE$, that was a typo; I've fixed it. Sep 21, 2017 at 15:06

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!

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...

• Those equations for G,H,I and J have an infinite number of solutions.
– ffao
Sep 20, 2017 at 23:39
• @ffao Well, if nothing else, this shows that it may be impossible without an actual calculation of the angles. Indeed there are multiple "valid" solutions for the equation set, making it unusable for this setup. I've updated my answer to reflect that. Thanks for the catch! Sep 21, 2017 at 0:03
• @ffao only one of the solutions for G, H, I and J is actually valid, as the construction allows for no variation in those angles. Sep 21, 2017 at 8:38
• @micsthepick In general I agree, especially because someone else has proven those numbers to work, but it's true that MY METHOD did not provide any verification that it HAD to be that solution. In reality, of course, only those angles fit the true geometry of the figure, but nothing proved that in my response. Sep 21, 2017 at 15:30

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?

• Sep 20, 2017 at 16:33
• Missed in this diagram is the length of ab is 10 @paparazzi. Sep 20, 2017 at 17:07
• @MeaCulpaNay Length and angle are not the same Sep 20, 2017 at 20:51

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

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

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