Jake's Game of Questionable Quality [6/5+]

[Felissan]

/in

[Felissan]

VOTE: Jake The Wolfie

[Felissan]

often, but don't spam.

[Felissan]

Dibs!

Take the parachute

[Felissan]

Original Roll String: 1d6

1 6-Sided Dice: (1) = 1

[Felissan]

I attempt to assemble an egg!

[Felissan]

Check inventory, while lovingly incubating the egg

[Felissan]

Poke the PokeCube™

[Felissan]

Construct the square root of 14 using the compass and ruler on a piece of paper by following the instructions given below and keep the construction lines:

Spoiler: Construction Instructions

I highly encourage you to actually try this using a tool like geogebra.

Part I: Constructing the square root of 2.
1: Start with a line segment AB and extend it to be a line.
2: Draw a circle with center A and radius AB. Label the point D as the other point on the circle that intersects the line.
3: Draw a circle with center D and radius BD.
4: Draw a circle with center B and radius BD.
5: Label the two intersection points from steps 4 and 5 as E and F.
6: Connect the line segment EF. Label one of the intersection point on EF with the circle from step 3 as C. Notice that steps 3-6 constructed a line EF that is perpendicular to the line AB.
7: Connect points B and C to form the segment BC. By the Pythagorean Theorem, the hypotenuse of triangle ABC has length equal to the square root of 2 (given the segment AB has length 1).

Part II: Constructing the square root of 3 from the square root of 2.
8: Extend the line segment BC to be a line.
9: Using a process similar to steps 2-6 above, create the line perpendicular to the line BC that intersects BC at the point B. Instructions 9a through 9d give a reminder of the process.
9a: Draw a circle with center B and radius BC. Label the point that intersects the line BC as G.
9b: Draw a circle with center C and radius CG.
9c: Draw a circle with center G and radius CG.
9d: Draw the line segment that intersects the two circles. This line is perpendicular to BC and intersects the point B.
10: Using a compass, measure the length AB. Then draw a circle with radius AB and center B.
11: Mark the point where the circle from step 10 intersects the line segment from step 9 as B2. The line segment should have length AB.
12: Connect points B2 and C to form triangle CBB2. By the Pythagorean Theorem, the line CB2 has length proportional to the square root of 3.

Part III: Constructing the square root of n in general:

13: Construct the square root of n - 1. Note that the point C should be shared by the hypotenuses of each triangle formed up to this point. The vertex of the right angle in the last triangle that constructs the square root of n - 1 is B_{n - 3} and the remaining vertex is B_{n - 2}. (Note that point B should be considered point B1.)
14: Construct the line perpendicular to CB_{n - 2} that intersects the point B_{n - 2}. Refer to steps 2-6 or 9 for a reminder of the process.
15: Use the compass to measure the length AB. Draw a circle with radius AB and center B_{n - 2}.
16: Label the point where the circle from step 15 intersects the line from step 14 as the point B_{n - 1}.
17: Connect points C and B_{n - 1}. By the Pythagorean Theorem, this line segment has length proportional to the square root of n.

Sell it as a piece of art.

Oh, you thought you could get away with not showing the picture, didn't you?

Spoiler:

And you were right, because I made it instead.

[Felissan]

Re-check my inventory!

[Felissan]

Check inventory again, now that I can actually read it

[Felissan]

Gently poke the wolf pupper with the PokeCube™

[Felissan]

Throw the PokeCube™ in StrangerCoug's direction. I choose you, pupper!

[Felissan]

Original Roll String: 1d4

1 4-Sided Dice: (4) = 4

[Felissan]

Eat Ircher's king-side bishop

[Felissan]

Look for egg pieces

StrangerCoug has started the weekly egg piece hunt! Anyone may join, and they are encouraged to do so.

/in

[Felissan]

Walk around in tall grass

[Felissan]

Join Ircher