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Alice and Bob are looking at each other, both turn $10$ degrees and now they both can directly see Claire. If they continue turning in their same directions before, Alice will directly able to see Eric after turning $10$ degrees more, and Bob will directly able to see David after turning $20$ degrees more. But Alice cannot see David because Claire blocks the view, and Bob cannot see Eric because David blocks the view.

If the distance between Claire and Eric is $5$ meters, What is the distance between David and Eric?

enter image description here

This image is not accurate though.

Note: No computer tag is for solving this question by hand. You can draw the diagram with a computer.

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  • $\begingroup$ This is so confusing. $\endgroup$ – paparazzo Sep 20 '17 at 11:00
  • $\begingroup$ Could you just show a diagram? $\endgroup$ – paparazzo Sep 20 '17 at 11:46
  • $\begingroup$ The answer still requires a calculator to determine, which would seem to contradict the no-computers tag. $\endgroup$ – micsthepick Sep 20 '17 at 11:50
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    $\begingroup$ @Seyed how did you arrive at that? $\endgroup$ – Irishpanda Sep 20 '17 at 20:56
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    $\begingroup$ I suspected that may end up being the case, I've been trying to show that geometrically myself for hours :\ $\endgroup$ – Irishpanda Sep 20 '17 at 22:27
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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.

enter image description here

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.

enter image description here

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:

enter image description here

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:

enter image description here

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:

enter image description here

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 $\triangle CED$:

enter image description here

$\left | CE \right |=\left | ED \right |$ 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.

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  • $\begingroup$ I cannot add more than two pictures because of rep... $\endgroup$ – sila Sep 21 '17 at 12:43
  • $\begingroup$ 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) $\endgroup$ – Rubio Sep 21 '17 at 13:55
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    $\begingroup$ @Rubio FE = AF = AC = CB = CF. Each equality is true because the two sides involved form an isosceles triangle. $\endgroup$ – Mike Earnest Sep 21 '17 at 14:01
  • $\begingroup$ @Rubio I edited that part a bit, sorry for jumping that conclusion so fast. $\endgroup$ – sila Sep 21 '17 at 14:14
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    $\begingroup$ Great solution, welcome to PSE! $\endgroup$ – Ankoganit Sep 21 '17 at 15:16
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The answer is:

$5$ meters.


The actual to-scale figure is something like this:

enter image description here

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

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  • $\begingroup$ how does canceling 2's imply $\sin30\sin20=\sin10\sin80$? $\endgroup$ – JMP Sep 21 '17 at 11:35
  • $\begingroup$ what's $\angle DBE$? $\endgroup$ – JMP Sep 21 '17 at 11:46
  • $\begingroup$ @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. $\endgroup$ – Ankoganit Sep 21 '17 at 15:06
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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:

enter image description here

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

enter image description here

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.

enter image description here

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

enter image description here

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:

enter image description here

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

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    $\begingroup$ Those equations for G,H,I and J have an infinite number of solutions. $\endgroup$ – ffao Sep 20 '17 at 23:39
  • $\begingroup$ @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! $\endgroup$ – Mister B Sep 21 '17 at 0:03
  • $\begingroup$ @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. $\endgroup$ – micsthepick Sep 21 '17 at 8:38
  • $\begingroup$ @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. $\endgroup$ – Mister B Sep 21 '17 at 15:30
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enter image description here

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?

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(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 $t_1$.

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  • $\begingroup$ where is the answer? $\endgroup$ – Oray Sep 21 '17 at 17:36
  • $\begingroup$ This problem have an infinity of solutions. $\endgroup$ – Kerkouch Sep 21 '17 at 17:40
  • $\begingroup$ Just replace t_1 with any value in ]10, 140[ and you will get the right value for the distance David and Eric. $\endgroup$ – Kerkouch Sep 21 '17 at 17:42
  • $\begingroup$ If t_1=90 degrees then, the distance will be 7.79 meters. $\endgroup$ – Kerkouch Sep 21 '17 at 17:43
  • $\begingroup$ you are not right actually. t1 is in between 10 to 140, that's right, but that doesnt mean it can be any value between those. I understand your dilemma but you are not including many other conditions, such as you are ignoring the angles at the beginning 10 degrees, you are just adding their sum and equal to that t1. you assume that the intersection point C just a random point. $\endgroup$ – Oray Sep 21 '17 at 17:48

protected by Rand al'Thor Sep 30 '17 at 10:09

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