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Puzzles

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KIAman said:
I applaud you for failing the reading comprehension test. Or... it's possible you never heard of FB, but using occam's razor, it's simpler to assume my initial assumption.
Click to expand...

Ok I have one for you.

What year long event ended in 1635?
 
Cozarkian said:
You can't logically conclude that one paw represents half of the number represented by two paws. Thus, the answer is

One paw = (? - 5)/10
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The zoomed in photo is worth half that of a zoomed out photo, so therefore, one paw must be half of two paws. Or something like that.
 
23435289_1753156528050946_3064705066079180931_n.jpg
 
the bottom 2 angles are 80 degrees (20+60 & 70+10), which means the top angle is 0 degrees (160-80-80) which means it's not a valid triangle
 
I can see how to solve it but that would involve remembering trig and I'd rather not.

1) Fill in all the easy angles.
2) assign an arbitrary length to the bottom side (1 would work nicely)
3) Solve for the length of the two other sides of the outside triangle using trig that I don't want to remember
4) Solve for the length of the right side of the 60-80-40 triangle which also yields the length of the right side of the top triangle
5) Solve the rest of the top triagle
6) Solve for x
 
IronWing said:
I can see how to solve it but that would involve remembering trig and I'd ratther not.
Click to expand...
I thought it'd take longer than it did, it didn't. I didn't use any formulas or anything, just logic'd out 180 degrees for each triangle. I imagine a trig teacher would thwack me in the back of the head for not showing my work though.
 
[DHT]Osiris said:
I thought it'd take longer than it did, it didn't. I didn't use any formulas or anything, just logic'd out 180 degrees for each triangle. I imagine a trig teacher would thwack me in the back of the head for not showing my work though.
Click to expand...
I don't think you can. You end up with two angles that are unconstrained without using side lengths. I'll look again to be sure.

One can solve the big triangle, no problem
One can solve the lower three triangles
One can solve for the angle at the bottom of the triangle containing angle x
One can constrain angle x + the angle above x to 150 degrees
One can constrain the angle across from x + the angle above it to 140 degrees

I think that's it for simple visual inspection.
 
IronWing said:
I don't think you can. You end up with two angles that are unrestrained without using side lengths. I'll look again to be sure.
Click to expand...
You get to a point where you've got two angles to figure out inside the primary triangle where x is, x and one more I labeled y, you also have the angles for complete triangles outside of those angles, so you can work out two formula where x and y need to fit to make 180 degree triangles between them (as well as the triangle housing x itself), only one pair of numbers works for all 3 equations.

Again, I'm sure there's a better way to do that where a smarter person than me has already created a formula for, I just slogged through it.
 
Okay, using your method I have four equations and four variables so we should be good.

x + x' = 150
y + y' = 140
x' + y' = 160
x + y = 130

x = 30
x' = 120
y = 100
y' = 40

works 🙂
 
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