# The Ultimate Battle – Episode 2 – On the edge

• Nick: Grigori
Server: ts1.travian.com

Arrange the logs in "T" form and fix them with one edge touching the exit and two touching the corners.

## Images

• he should use both logs, to form 90 degree . the logs will fit exactly in the ditch and he can walk on one of the logs to arrive att he door . nick unclejoe server comx

• Assuming the logs have some width (not necessary though), here's the math

Acidburn on com1

• com83(beta): rastogi2008

com29(NYS5x): Slavefactory

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It plainly marques, four my revue, miss steaks eye kin knot sea.
Eye strike a quay and type a word And weight four it two say
weather eye am wrong oar write It shows me strait a weigh.

• our sticks are too short. So we need to find a better way. we understand that if we put one of the sticks over the corner we could possible create a T-shaped bridge and pass over. To pass from the edge of the inner square to the corner of the outer platform. This is pythagoras theorem: Square root of 300^2 + 300^2 = 425. This means we need, 425 - 300 = 125, extra cm from our T-shape.

So you put one stick in the corner of the outer rim. This creates a triangle in the corner which can be broken up into 2 triangles with the angles 90,45 and 45. This means that the new point we can put our 2nd stick on is half the length of the first stick times Tan(45) which is equal to 150. This means if we put 150 together with our 300 cm stick we have 450 cm of bridge when we only need 425. Ergo we get to safety.

account: Pottan

Server: TX3

• well I hope the guy has a belt and / or a pair of socks so he can tie the two poles together and make it a little bit longer so that it does reach across the gap and a bit stronger also, then crawl across

Muger com3

• Solution in attachment :

Explaination

keep 1st end at top edge and 2nd end at right edge,

make the angle to both edges 45 degrees

keep 1st end on middle of 1st ladder, and 2nd end on the top-right corner of exit-area

Math Proof :

DY = XY = 300

lets say sq2 = square root of (2) = 1.414

DX = 300 * sq2 = DC + CX

DC = 300

AB = 300, also AC = BC and AC + BC = 300

hence AC = 150

also angle CAX = 45 degrees = angle AXC

angle XCA = 90 degrees

hence AC = CX = 150

DX = DC + CX = 300 + 150 = 300 * (1 + 0.5) = 300 * (1.5)

length needed is 300 * sq2 : 300 * 1.414

hence ladder provide more lenght and are feasible

Domain : COM 1

• Picture attached

## Images

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