Obligatory AMSP Week 1 Post
by djmathman, Jul 8, 2013, 1:16 AM
Darn, AMSP is hard.
Even though it's only been a week I've learned a good deal, including:
Here are some of the problems I've conquered on the worksheets and tests this week. If anybody knows the sources to these problems, let me know!
Even though it's only been a week I've learned a good deal, including:
- Don't let Nicky Sun steal your name tag on the first day; if he does, threaten him with push-ups later for an easy win
- Cosmin is too OP
Here are some of the problems I've conquered on the worksheets and tests this week. If anybody knows the sources to these problems, let me know!
Quote:
Let 
be the number of 
-tuples 
with values in 
and 
for which the sum 
is odd, and 
the number of -tuples for which the sum is even. Prove that
![\[\dfrac{O_n}{E_n}=\dfrac{2^n-1}{2^n+1}.\]](//latex.artofproblemsolving.com/0/6/1/06122299edef254ebcd50b02ddcddf246e04996a.png)
be the number of 
-tuples 
with values in 
and 
for which the sum 
is odd, and 
the number of -tuples for which the sum is even. Prove that![\[\dfrac{O_n}{E_n}=\dfrac{2^n-1}{2^n+1}.\]](http://latex.artofproblemsolving.com/0/6/1/06122299edef254ebcd50b02ddcddf246e04996a.png)
![\[\sum_{k\leq n, 2\nmid k}\dbinom nk 3^{n-k}=2^{2n-1}-2^{n-1}.\]](http://latex.artofproblemsolving.com/6/e/6/6e6e9f9e02a65ceb1125c534f31d4cbf6ec51fe1.png)
and 
.
odd) will remain (and in fact be counted twice). So letting ![\[2X=2^{2n}-2^n\implies X=2^{2n-1}-2^{n-1}.\qquad\blacksquare\]](http://latex.artofproblemsolving.com/9/f/3/9f3e99af64d22e5c16dbd6def02c3b8a6ca835bf.png)
,
must be equal to one and the rest must be equal to zero. If any 
that satisfies the conditions 
.
.
.
,
.
and our desired trivially follows. 
Quote:
Let 
be any positive integer. Show that ![\[\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\dbinom nk=1+\dfrac12+\ldots+\dfrac1n.\]](//latex.artofproblemsolving.com/e/3/2/e32b682c54178b3583ba6375505321852e5a39bb.png)
be any positive integer. Show that ![\[\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\dbinom nk=1+\dfrac12+\ldots+\dfrac1n.\]](http://latex.artofproblemsolving.com/e/3/2/e32b682c54178b3583ba6375505321852e5a39bb.png)
,![\[\sum_{k=1}^{n+1}\dfrac{(-1)^{k-1}}k\dbinom {n+1}k=\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\dbinom nk+\dfrac1{n+1}.\]](http://latex.artofproblemsolving.com/3/4/2/342a2e2b5b0fec84047cb76c65ff359eff7d6cf9.png)
Splitting up the sum on the left and simplifying, we have![\begin{align*}\sum_{k=1}^{n}\dfrac{(-1)^{k-1}}k\dbinom {n+1}k+\dfrac{(-1)^n}{n+1}&=\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\dbinom nk+\dfrac1{n+1}
\\\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\left[\dbinom{n+1}k-\dbinom nk\right]&=\dfrac{1-(-1)^n}{n+1}
\\\sum_{k=1}^n\dfrac{(-1)^{k-1}}k\dbinom n{k-1}&=\dfrac{1-(-1)^n}{n+1}.\end{align*}](http://latex.artofproblemsolving.com/f/b/8/fb830a057f884af1b5f9ee443251ef58c8a6a940.png)
Now multiplying both sides by 
and simplifying gives
Now it is not hard to show that this is true. In the interest of time, I will say that it is easy to move everything to one side and condense using Justin Stevens to show that it is true. Quote:
We are given 
intervals on the real line. Prove that there either exist 
intervals which are pairwise disjoint or intervals with nonempty intersection.
intervals on the real line. Prove that there either exist 
intervals which are pairwise disjoint or intervals with nonempty intersection.
.
along these intervals with the equivalence relation ![\[A_i<A_j\text{ if } \min(A_i)<\max(A_j)\text{ and }A_i\cap A_j\neq\emptyset.\]](http://latex.artofproblemsolving.com/9/9/2/9926c0113dcbddd6ec365a165bf0a5a744fc4c5d.png)
In this case, antichains are formed by pairwise disjoint intervals and chains are formed by intervals that are not pairwise disjoint. I shall first prove that there are either eight pairwise disjoint intervals or eight intervals which pairwise have nonempty intersection. Suppose not. Then the maximum length of both the chains and the antichains are both 
.
,
,
,
,
,
.
.
passes through each of the other seven intervals, and we are done. Quote:
Points 
are chosen on the sides 
respectively of a triangle 
. Denote by 
the centroids of triangles 
, 
, 
respectively. Prove that the lines 
, 
, 
are concurrent if and only if lines 
are concurrent.
are chosen on the sides 
respectively of a triangle 
. Denote by 
the centroids of triangles 
, 
, 
respectively. Prove that the lines 
, 
, 
are concurrent if and only if lines 
are concurrent.
be the midpoints of 
,
,
respectively. It is clear that we no longer have to focus on the centroids but rather these midpoints. Note that by the Ratio Lemma,![\[\dfrac{M_aC_1}{M_aB_1}=\dfrac{AB_1\sin\angle M_aAB}{AC_1\sin\angle M_aAC}=1.\]](http://latex.artofproblemsolving.com/2/d/1/2d13074917cc5c45beaea48dbf940f4dd68cc7b9.png)
We can derive similar equivalences concerning 
and 
.![\[\left(\dfrac{AB_1}{AC_1}\cdot\dfrac{BC_1}{BA_1}\cdot\dfrac{CA_1}{CB_1}\right)\left(\dfrac{\sin\angle M_aAB_1}{\sin\angle M_aAC_1}\cdot\dfrac{\sin\angle M_bBC_1}{\sin\angle M_bBA_1}\cdot\dfrac{\sin\angle M_cCA_1}{\sin\angle M_cCB_1}\right)=1.\]](http://latex.artofproblemsolving.com/e/7/e/e7ed51791e1d7a404bcb3063d2c2210e224d4185.png)
Now it is clear that normal Ceva for 
,Quote:
Let be an isosceles triangle with 
. Its incircle touches 
in 
and 
in 
. A line distinct of 
goes through 
and intersects the incircle in 
and 
. Line intersects lines 
ad in 
and 
. Prove that 
.
be an isosceles triangle with 
. Its incircle touches 
in 
and 
in 
. A line distinct of 
goes through 
and intersects the incircle in 
and 
. Line intersects lines 
ad in 
and 
. Prove that 
.
.
w.r.t line 
gives![\[\dfrac{BE}{EX}\cdot\dfrac{XG}{GA}\cdot\dfrac{AL}{LB}=1.\]](http://latex.artofproblemsolving.com/0/0/3/003e6f8c9c8c0a11eba7ffcfc0f4e87c6d293339.png)
Similarly, applying Menelaus on 
givese![\[\dfrac{BE}{EX}\cdot\dfrac{XF}{FA}\cdot\dfrac{AK}{KB}=1.\]](http://latex.artofproblemsolving.com/e/5/3/e53726a41921d4626630812a93f76f16eee9b0f0.png)
Multiplying these together and rearranging gives![\[\left(\dfrac{BE^2}{GA\cdot FA}\right)\left(\dfrac{XG\cdot XF}{EX^2}\right)\left(\dfrac{AL\cdot AK}{BL\cdot BK}\right)=1.\]](http://latex.artofproblemsolving.com/4/5/e/45e05fe6ca204313251fe4f514fcdb48607f3440.png)
Note that by PoP we have 
and 
.
Rearranging gives
Adding 
to this equality gives 
,Quote:
Let be a triangle and let 
and 
be the squares erected on the sides which are directed towards the exterior of the triangle. Let 
be the circumcenter of triangle 
. Prov that 
is the -symmedian of triangle .
be a triangle and let 
and 
be the squares erected on the sides which are directed towards the exterior of the triangle. Let 
be the circumcenter of triangle 
. Prov that 
is the -symmedian of triangle .
.
onto 
be 
and 
be 
respectively. Note that since 
and 
,
,
.
.![\[\dfrac{YP}{YQ}=\dfrac{YP/YA}{YQ/YA}=\dfrac{AD/AX}{AE/AX}=\dfrac{AD}{AE}=\dfrac{AT}{AV}=\dfrac{AB}{BC}.\]](http://latex.artofproblemsolving.com/e/e/d/eedc8d28b43f64082683007628ca1e025632e6b3.png)
Since the distances from 
,This post has been edited 6 times. Last edited by djmathman, May 31, 2020, 9:58 PM
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