You are Here: Home >< Maths

# The Proof is Trivial! Watch

1. Problem 511

Show that

2. Solution 511
Spoiler:
Show

Let
Therefore
Therefore

Let
Therefore

And hence

Therefore .
3. (Original post by Indeterminate)
Problem 511

Show that

Solution 511

Quite like this one.

Let be our sum.
Then
Then

edit: aw, i was too slow.
4. (Original post by Renzhi10122)
Solution 511

Quite like this one.
edit: aw, i was too slow.
Haha your solution is much nicer than my "brute force" though.
5. (Original post by 16Characters....)
Haha your solution is much nicer than my "brute force" though.
Well, it's pretty much exactly the same tbh.
6. (Original post by Renzhi10122)
Well, it's pretty much exactly the same tbh.
Thought I might be the first to solve one of these for once... two people get in before me

Anyhow my solution was similar, but naturally more messy, that the two that have gone before.

Posted from TSR Mobile
7. (Original post by Krollo)
Thought I might be the first to solve one of these for once... two people get in before me

Anyhow my solution was similar, but naturally more messy, that the two that have gone before.

Posted from TSR Mobile
My question stil remains
8. Problem 512

Prove that in a complete bipartite planar graph connecting two sets; one of 3 elements and 1 of 2 elements, one of the nodes in the graph always lies geographically inside of the cycle formed by all of the other nodes.

Hence deduce that the "Three Utilities Problem" is impossible. (https://en.wikipedia.org/wiki/Three_utilities_problem )
9. Nice, nice, this thread seems to be picking up, I might try crafting one or two questions later tonight if I find the time.
10. (Original post by 16Characters....)
Problem 512

Prove that in a complete bipartite planar graph connecting two sets; one of 3 elements and 1 of 2 elements, one of the nodes in the graph always lies geographically inside of the cycle formed by all of the other nodes.

Hence deduce that the "Three Utilities Problem" is impossible. (https://en.wikipedia.org/wiki/Three_utilities_problem )
Solution 512

Consider any face. Then any two adjacent vertices will be in different sets. It follows that the face is composed of an even number of vertices, namely 4. Therefore, the outer face is comprised of 4 vertices, leaving one vertex in the area not reached by this face, i.e. the area insides the cycle made by these 4 vertices. In particular, this vertex is part of the set of 3 vertices, since the outer face is comprised of 2 vertices from each set.

In the three utilities problem, it is clear that the outer face cannot have 6 vertices. If it did then this would lead to a contradiction immediately due to crossing lines within ths cycle formed. Therefore the outer face has 4 vertices, so 2 vertices lie inside the cycle formed. We note now that this graph can be formed by adding one vertex inside the outer cycle to the graph formed previously. However, the vertex inside this cycle splits the face into two faces, one of which our added vertex resides in. It must be joined by edges to the vertex that is in the interior of the cycle and the 2 vertices that the interoir vertex is not connected to, but this is impossible.
11. Problem 513

Stupid Bob claimed:

In each case, find all values of x and y such that Stupid Bob is serendipitously correct.
12. (Original post by Krollo)
Problem 513

Stupid Bob claimed:

In each case, find all values of x and y such that Stupid Bob is serendipitously correct.
Spoiler:
Show
Would it just be something like solving for the first one, and similarly for the rest?
13. (Original post by Krollo)
Problem 513

Stupid Bob claimed:

In each case, find all values of x and y such that Stupid Bob is serendipitously correct.
so for individual cases in isolation , or all cases correct simultaneously?
14. (Original post by GCSEHELPPLS)
so for individual cases in isolation , or all cases correct simultaneously?
In isolation, I reckon.

Posted from TSR Mobile
15. (Original post by Krollo)
In isolation, I reckon.

Posted from TSR Mobile
k thanks, I've only done the first one so far, is it worth submitting?
16. (Original post by GCSEHELPPLS)
k thanks, I've only done the first one so far, is it worth submitting?
Sure, go for it!
17. Spoiler:
Show
sin(x+y)-sin(x)-sin(y) = 0
sin(x)cos(y)+cos(x) sin(y) - sin(x)-sin(y)=0
sin(x) (cos(y)-1)+(cos(x)-1) sin(y) = 0

subbing double angle formulas in for x/2 and y/2 gives
2sin(x/2)cos(x/2)(cos^2(y/2)-sin^(y/2)-1)+2sin(y/2)cos(y/2)(cos^2(x/2)-sin^(x/2)-1)= 0

1=sin^2(a)+cos^2(a)so we get
2sin(x/2)cos(x/2)(-2sin^2(y) + 2sin(y/2)cos(y/2)(-2sin^2(x)=0
-4 sin(x/2) sin(y/2) (sin(x/2)cos(y/2)+cos(x/2) sin(y/2)= 0
-4 sin(x/2) sin(y/2) sin(x/2+y/2) = 0

so x+y=2pi or
sinx/2 =0 or
siny/2=0

for x and y being above than or equal to 0 and less than 2 pi, this gives x=2pi-y but if x or y are 0,
the other one can be anything
18. sorry very messy - i don't know how to use latex
19. (Original post by Zacken)
Sure, go for it!
I've submitted in plain text, just asking how do u get latex on the post ? thanks
20. (Original post by GCSEHELPPLS)
I've submitted in plain text, just asking how do u get latex on the post ? thanks
Here you go. This is quite helpful:

http://www.thestudentroom.co.uk/wiki/LaTex

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 11, 2017
Today on TSR

### Am I pregnant?

...or just paranoid?

### Top study tips for over the Christmas holidays

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.