15
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Lieutenant, launch the polar probes!

Launching probes, sir. Failure, sir. Probes did not land at the poles.

Well!? Where are they?

Unknown, sir. Somewhere off-axis. They did land on opposite sides of the planet.

Launch another pair!

Another failure, sir. Same thing. They are opposite each other, but not at the poles.

Same places as the first pair?

No, sir. Completely different.

Dammit, find them. Tachyon ranging pulse!

All probes responding. Linear distances from ship are 7, 9, 11, and [CENSORED] SDUs*.

(new voice) HAHAHAHAHA

Ensign! What is so funny?

Sir, at our current altitude, the area of the planet's surface we can see is exactly equal to the area of a flat circle of [CENSORED] radius! Funny coincidence, sir.

Shut up, Wesley!

*SDU : Standard Distance Unit

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  • 4
    $\begingroup$ Is the implication that both censored numbers are the same? $\endgroup$ – Mike Earnest Apr 23 '15 at 0:06
  • $\begingroup$ I'm trying to decide if it's Wesley Crusher or if "Sir" is the robot from Clone High. $\endgroup$ – Ian MacDonald Apr 23 '15 at 1:20
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    $\begingroup$ @MikeEarnest, yes, the number being censored happens to be an unfathomable insult in Galactic Common, and for the sake of peace, it is always filtered by the Universal Translator $\endgroup$ – Anon Apr 23 '15 at 11:54
  • $\begingroup$ Bet they are making an 455 of themselves $\endgroup$ – Andrew Smith Apr 23 '15 at 16:26
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    $\begingroup$ @EngineerToast,@Kevin "[CENSORED]" was merely said last due to the natural reluctance of anyone to utter such unspeakable words. The other distances are also listed in no particular order. $\endgroup$ – Anon Apr 23 '15 at 20:49
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The censored number is

3

Here is a cross section of the planet containing all four probes, illustrating some terms:

$\qquad\qquad\qquad\qquad$enter image description here

To clarify, the distances to the four probes are $D_{A^+},D_{A^-},D_{B^+},$ and $D_{B^-}$, and these four are some rearrangement of the numbers $7,9,11$, and [CENSORED]. Without loss of generality, assume $$ D_{A^+}=[CENSORED] $$ Now, applying Law of Cosines to the triangle with angle $\varphi$, we get $$ D_{A^-}^2=H^2+R^2-2RH\cos\varphi $$ and in the same vein, $$ D_{A^+}^2=H^2+R^2-2RH\cos(180-\varphi)=H^2+R^2+2RH\cos\varphi. $$ Adding these together, $$ D_{A^-}^2+D_{A^+}^2=2R^2+2H^2\tag{1} $$ Similarly, $$ D_{B^-}^2+D_{B^+}^2=2R^2+2H^2\tag2 $$ Combining (1) and (2), we get that $D_{A^+}^2=D_{B^+}^2+D_{B^-}^2-D_{A^-}^2$, so the three possibilities for $D_{A^+}^2$ are $$ 11^2+9^2-7^2=153\qquad 7^2+11^2-9^2=89\qquad 9^2+7^2-11^2=9 $$ The left two cases would imply that $D_{A^+}$ was one of the larger two distances among $7,9,11$ and [CENSORED] (since $\sqrt{153}>\sqrt{89}>9$).

What are we to make of the Wesley's statement? The area they can see is at most half the planet's area, so at most $2\pi R^2$. Since Wesley said this is equal to $\pi\times[CENSORED]^2$, it follows that $\pi\times[CENSORED]^2\le 2\pi R^2$, so $$ [CENSORED]^2=D_{A^+}^2\le 2R^2\le R^2+H^2\tag 3 $$ since $R\le H$. Combining (1) and (3), we see that $$D_{A^-}^2=2R^2+2H^2-D_{A^+}^2\ge R^2+H^2\ge D_{A^+}^2$$ so that $D_{A^+}$ is the smaller of the two $A$ probe distances. This means that the above bold statement cannot be true, so that we must have $D_{A^+}^2=9$, giving us the answer as claimed.

We can also make an educated guess as to why this number is so vulgar:

it looks like a sideways butt.

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  • 1
    $\begingroup$ Good job! But I demand you apologize for the repeated swearing in this answer. $\endgroup$ – Lopsy Apr 24 '15 at 9:14
  • $\begingroup$ Nice :-) The reason I'm proud of this puzzle is because the radius of the planet, the altitude of the ship, and the distances between the probes are all still indeterminate, but [CENSORED] is always the same. $\endgroup$ – Anon Apr 24 '15 at 21:54
  • $\begingroup$ @Anon Yeah, this is pretty satisfying! Is this one of your creation? $\endgroup$ – Mike Earnest Apr 25 '15 at 20:15
  • $\begingroup$ Yes, it's my creation. Glad you enjoyed it. $\endgroup$ – Anon Apr 27 '15 at 17:01
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PARTIAL ANSWER

This how far I've gotten. I hope someone with more a recent education than I can take correct what I have wrong and finish it up.


tl;dr

Here are pieces I think will help: $$D_A^2=49=r_P^2+z_S^2-2z_Sz_A$$ $$D_B^2=81=r_P^2+z_S^2-2z_Sz_B$$ $$D_C^2=121=r_P^2+z_S^2-2z_Sz_C$$ $$D_D^2=??=r_P^2+z_S^2-2z_Sz_D$$ $$5z_A-9z_B+4z_C=0$$ $$\frac{1}{z_S}=\frac{z_A-z_B}{4}=\frac{z_A-z_C}{9}=\frac{z_B-z_C}{5}$$ $$r_C^2=2r_Ph$$ $$r_P^2=\pm(x_Ax_B+y_Ay_B+z_Az_B)$$ $$r_P^2=\pm(x_Cx_D+y_Cy_D+z_Cz_D)$$


Name the 4 points A, B, C, D and their coordinates as $(x_A,y_A,z_A)$, etc.
Name the ship S and its coordinates are $(x_S,y_S,z_S)$
Name the planet P and its coordinates are $(x_P,y_P,z_P)$ with a radius $r_P$

For convenience, set the center of the planet $P$ to be at the origin.
In addition, orient the axes such that the ship $S$ is directly above the planet along the z-axis.

This tells us that $x_P=y_P=z_P=x_S=y_S=0$

The general equation for the distance between two points in 3D space is $$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$ Therefore, the distance between the ship $S$ and point $A$ is $$D_A=7=\sqrt{(x_S-x_A)^2+(y_S-y_A)^2+(z_S-z_A)^2}$$ $$D_A=7=\sqrt{(0-x_A)^2+(0-y_A)^2+(z_S-z_A)^2}$$ $$D_A=7=\sqrt{x_A^2+y_A^2+(z_S^2-2z_Sz_A+z_A^2)}$$ $$D_A=7=\sqrt{(x_A^2+y_A^2+z_A^2)+(z_S^2-2z_Sz_A)}$$

Now, since the planet's center is at the origin, we know $x_A^2+y_A^2+z_A^2=r_P^2$
We can plug that in and square both sides to get $$D_A^2=49=r_P^2+z_S^2-2z_Sz_A$$ It is trivial to expand that to the other 3 points $$D_B^2=81=r_P^2+z_S^2-2z_Sz_B$$ $$D_C^2=121=r_P^2+z_S^2-2z_Sz_C$$ $$D_D^2=??=r_P^2+z_S^2-2z_Sz_D$$

Right now, we're sitting at 4 equations and 7 unknowns ($r_P,z_S,z_A,z_B,z_C,z_D,D_D$). That's close but we're not there yet. We can play around with the formulas for $D_A$ and $D_B$ because they're values are known. $$D_A^2+32 = 49+32=81=D_B^2$$ $$r_P^2+z_S^2-2z_Sz_A+32=r_P^2+z_S^2-2z_Sz_B$$ $$-2z_Sz_A+32=-2z_Sz_B$$ $$2z_Sz_B+32=2z_Sz_A$$ $$z_Sz_B+16=z_Sz_A$$ We can do the same thing for $D_A$ and $D_C$ as well as $D_B$ and $D_C$ $$z_Sz_C+36=z_Sz_A$$ $$z_Sz_C+20=z_Sz_B$$ If we solve all three for $z_S$, we can set them equal. $$z_S=16/(z_A-z_B)=36/(z_A-z_C)=20/(z_B-z_C)$$ $$z_S=4/(z_A-z_B)=9/(z_A-z_C)=5/(z_B-z_C)$$ $$\frac{1}{z_S}=\frac{z_A-z_B}{4}=\frac{z_A-z_C}{9}=\frac{z_B-z_C}{5}$$ This gives us the relationship $$5z_A-9z_B+4z_C=0$$


Let's look at the second observation. The formula for the surface area a spherical cap $A_P$ that is distance $h$ from the origin and the area of a circle $A_C$ are given by $$A_P=2\pi r_Ph$$ $$A_C=\pi r_C^2$$ We're told those are equal so we can drop the $\pi$ to get the equality $$r_C^2=2r_Ph$$

Unfortunately, this introduces only 1 new equation and 2 new unknowns. I think the solution will involve some trig to find $h$ based on $r_P$, $z_S$, and tangent lines but I don't have the brainpower for that right now.


I'm sure that the probes landing on opposite sides matters but I couldn't get it to do anything useful because it added more unknowns instead of removing them. Here's what I wrote for that section:

We also know that the probes landed on opposite sides of the sphere. That means the line connecting them passes through the center which means it is of length $2r_P$. That gives us the following for the distance between probe $A$ and $B$. (Here I have to assume that the first two distances are for the first two probes. If that's not the case, this section must be reworked.) $$2r_P=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2+(z_A-z_B)^2}$$ $$2r_P=\sqrt{(x_A^2-2x_Ax_B+x_B^2)+(y_A^2-2y_Ay_B+y_B^2)+(z_A^2-2z_Az_B+z_B^2)}$$ $$2r_P=\sqrt{(x_A^2+y_A^2+z_A^2)+(x_B^2+y_B^2+z_B^2)-2(x_Ax_B+y_Ay_B+z_Az_B)}$$ $$2r_P=\sqrt{(r_P^2)+(r_P^2)-2(x_Ax_B+y_Ay_B+z_Az_B)}$$ $$2r_P=\sqrt{2r_P^2-2(x_Ax_B+y_Ay_B+z_Az_B)}$$ $$4r_P^2=\pm2(r_P^2-(x_Ax_B+y_Ay_B+z_Az_B))$$ $$r_P^2=\pm(x_Ax_B+y_Ay_B+z_Az_B)$$

(I'm not super-confident in how the $\pm$ comes in once you square both sides but I think it has to be considered somewhere.)

We can do the same thing for the other two probes $$r_P^2=\pm(x_Cx_D+y_Cy_D+z_Cz_D)$$

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  • 1
    $\begingroup$ I can verify that this solution is way overcomplicating things. You're getting bogged down in algebra without a strategy for handling the equations. $\endgroup$ – Lopsy Apr 23 '15 at 21:05

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