/sci/ - Science & Math » Thread #11430277

9KiB, 605x381, prob_01.png

View Same Google iqdb SauceNAO

Anonymous Sun 01 Mar 07:44:16 2020 No.11430277 View Reply Original Report

Quoted By: >>11430283 >>11431673 >>11431700 >>11431916 >>11433381

ITT: /sci/ does a little contest math.

Post your immediate ideas for pic related. Don't wait for full solution if you can help it. Don't post if you know the answer. No cheating / Googling the answer. I will post another pic if we get the answer.

Anonymous

Anonymous Sun 01 Mar 2020 07:46:27 No.11430280 Report

Quoted By:

my first thought: it's not the 3-4-5 triangle because angles are ~ 36.87° and 53.13°.

Anonymous

Anonymous Sun 01 Mar 2020 07:46:57 No.11430283 Report

Quoted By: >>11430285

>>11430277
are we to assume this is a right angle triangle?

Anonymous

Anonymous Sun 01 Mar 2020 07:48:13 No.11430285 Report

Quoted By:

>>11430283
i don't see why we would. the problem does not specify a right triangle.

Anonymous

Anonymous Sun 01 Mar 2020 07:55:46 No.11430299 Report

Quoted By:

the fact that all three sides must be consecutive integers seems extremely forcing. it must be possible to simply write down the angles (using the length) and observe that one is double the other?

0 - 1 - 2 is obviously not a triangle.

1 - 2 - 3 triangle. how do we write down the angles? cosine law?

Anonymous

Anonymous Sun 01 Mar 2020 07:55:56 No.11430300 Report

Quoted By:

My first thought is split into cases for which angle is twice the other. Using the angle sum rule write all the angles as a function of one of the others. Use law of sin and cosine to setup 2 equations then solve.

Anonymous

Anonymous Sun 01 Mar 2020 07:59:29 No.11430308 Report

Quoted By: >>11430312 >>11431898

1 - 2 - 3 isn't a triangle
3 - 4 - 5 doesn't work as another anon pointed out
as you increase the side lengths the sides will get closer to being equal, so I put 2 - 3 - 4 into a calculator and it fits the criteria. I'm a brainlet though so I don't know how to prove it's the only possible solution with math

Anonymous

Anonymous Sun 01 Mar 2020 08:01:18 No.11430312 Report

Quoted By: >>11430338

>>11430308
*4 - 5 - 6 is the solution, not 2 - 3 - 4

Anonymous

Anonymous Sun 01 Mar 2020 08:01:32 No.11430314 Report

Quoted By:

sine law seems like a good chance here

a / sin A = b / sin B = c / sin C

becomes

s / sin A = s+1 / sin B = s+2 / sin C

and then do we know which one should be the double angle?

Anonymous

Anonymous Sun 01 Mar 2020 08:08:36 No.11430324 Report

Quoted By:

if n is big, then the triangle with sides n, n+1, n+2 has almost equal sides, so also almost equal angles, make that precise and then examine the finite number of remaining triangles

Anonymous

Anonymous Sun 01 Mar 2020 08:17:16 No.11430338 Report

Quoted By:

>>11430312

by the cosine law:

c^2 = a^2 + b^2 - 2ab cos(C)

36 = 16 + 25 - 40 cos(C)
1/8 = cos(C)

similarly

b^2 = a^2 + c^2 - 2ac cos(B)
25 = 16 + 36 - 48 cos(B)
27/48 = cos(B)

a^2 = b^2 + c^2 - 2bc cos(A)
16 = 25 + 36 - 60 cos(A)
45/60 = cos(A)
3/4 = cos(A)

hmm. taking the arccos is not particularly compelling ...

Anonymous

Anonymous Sun 01 Mar 2020 19:06:50 No.11431673 Report

Quoted By: >>11431678 >>11431685

>>11430277
Using a little linear algebra:
>We have the sum of 3 dot products is 180
>we have one dot product twice the other
>we have 3 magnitudes equal m0, m1, m2
>this puts dot product m1 and m2 dot products on the smaller angle
>we have now exhausted our unknowns with equations
>thus we don't even need to use the fact that m are all integers

Anonymous

Anonymous Sun 01 Mar 2020 19:08:02 No.11431678 Report

Quoted By:

>>11431673
>m1 and m2 on smaller angle
meant m1 and m0

Anonymous

Anonymous Sun 01 Mar 2020 19:09:47 No.11431685 Report

Quoted By:

>>11431673
I'm actually not entirely sure about that last part, but if we find an equation with multiple solutions, we find the one with consecutive integers, then plug in arbitrary consecutive numbers and prove those can't be integers.

Anonymous

Anonymous Sun 01 Mar 2020 19:13:38 No.11431700 Report

Quoted By:

>>11430277
Something along the lines of putting a vector of magnitude n, then another of magnitude n+2 and letting the magnitude of their substraction be n+1
Then use dot product to deduce the angle and ches ches wala wala winga

Anonymous

Anonymous Sun 01 Mar 2020 20:19:57 No.11431898 Report

Quoted By: >>11431900

>>11430308
>1 - 2 - 3 isn't a triangle
explain

Anonymous

Anonymous Sun 01 Mar 2020 20:21:28 No.11431900 Report

Quoted By:

>>11431898
nevermind, I'm retarded.

Anonymous

Anonymous Sun 01 Mar 2020 20:26:12 No.11431916 Report

Quoted By:

>>11430277
side lengths 3, 4, 5

Anonymous

Anonymous Mon 02 Mar 2020 07:13:44 No.11433381 Report

Quoted By:

>>11430277
Let ABC be with WLOG angles A=x, B=2x. Draw BD the angular bisector of B (D belongs to ac). Now BC is tangent to the circunsircle of ABD, since angles BAD=DBC. By the power of the point then CD*CA=BC*BC. Since BD is the angular bisector, AB/BC=AD/DC, and AC=AD+DC=(AB*DC/BC)+DC, therefore DC=AC/(1+AB/BC). Therefore BC^2=CD*CA=AC^2/(1+AB/BC).

Since angles A<B, so AC>BC. Therefore (AB, BC, AC) is one of the following: (x, x+1,x+2), (x+1, x, x+2), (x+2, x, x+1). Now substitute the triples into the equation: BC^2=AC^2/(1+AB/BC), and solve for x.

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches