## Search found 62 matches

Mon Nov 11, 2019 1:51 pm
Forum: Combinatorics
Topic: Vietnam tst 2004
Replies: 1
Views: 36998

### Re: Vietnam tst 2004

The equality holds only when $ABCDEF$ is also equiangular. Taking the $\triangle A_1BB_1$, it can be observed that when the perpendicular is dropped from $B$ on to $A_{1}B_{1}$, two right-angled triangles are formed, where the other two angles in each triangle are $30^\circ$ and $60^\circ$. Hence, w...
Tue Nov 05, 2019 1:48 am
Forum: Geometry
Topic: AIME II 2018 problem 4
Replies: 1
Views: 47355

### Re: AIME II 2018 problem 4

After the tedious calculations, I found the answers to be $a=25$ and $b=6$.
Therefore, $a+b=31$.

But I'm too tired to write the full solution right now. Hopefully, I will post it when I feel like! Till then, I invite someone else to prove my answer. Good luck! Tue Nov 05, 2019 1:01 am
Forum: Number Theory
Topic: Euler's Criterion
Replies: 3
Views: 52629

### Re: Euler's Criterion

Don't post such half-detailed problems, if you're really hoping for a solution to be posted.
Tue Nov 05, 2019 12:59 am
Forum: Number Theory
Topic: Euler's Criterion
Replies: 3
Views: 52629

### Re: Euler's Criterion

What are the values of $A$ and $N$?

Usually, $N$ is considered to be the symbol of "Natural Numbers".

But in this case, which Natural number?
Tue Nov 05, 2019 12:57 am
Forum: Combinatorics
Topic: Turkey TST 2014
Replies: 7
Views: 67402

### Re: Turkey TST 2014

BTW, I wonder...

This problem was posted over 5 years ago.

And no one posted a solution to this problem in half a decade!!! WOW!!!
Tue Nov 05, 2019 12:52 am
Forum: Combinatorics
Topic: Turkey TST 2014
Replies: 7
Views: 67402

### Re: Turkey TST 2014

Taking a $n\times n$ chessboard, where $n\equiv 2$ (mod $4$) and experimenting with smaller values of $n$ (like 6 and 10), yields the pattern which is denoted as series $B$ in the solution. From there, the rest was quite straight-forward! :D But yeah, it is actually an elegant problem, primarily bec...
Tue Nov 05, 2019 12:34 am
Forum: Combinatorics
Topic: Turkey TST 2014
Replies: 7
Views: 67402

The first step remains the same: cutting the board in half and denoting the upper-half as $BROWN$ and the lower-half as $GREEN$. Now, two brown worms ($B_1$ and $B_2$) start from the top-left corner; $B_1$ goes to the right and $B_2$ goes to the down direction. Upon reaching the end of the board, $B... Mon Nov 04, 2019 11:39 pm Forum: Combinatorics Topic: Turkey TST 2014 Replies: 7 Views: 67402 ### Re: Turkey TST 2014 The problem can be approached by first cutting the board in half; let the upper-half be called$BROWN$and the lower-half be called$GREEN$. Now, a pair of brown worms each from the top-left corner move to each of the squares (except the last one) in the diagonal of the LHS$1007\times1007$square ... Mon Nov 04, 2019 9:32 pm Forum: Combinatorics Topic: Turkey TST 2014 Replies: 7 Views: 67402 ### Re: Turkey TST 2014 The problem can be approached by first cutting the board in half; let the upper-half be called$BROWN$and the lower-half be called$GREEN$. Now, a pair of brown worms each from the top-left corner move to each of the squares (except the last one) in the diagonal of the LHS$1007\times1007\$ square o...
Thu Oct 31, 2019 9:11 pm