[/quote]
,
be a triangle with centroid 
.
and 
are chosen on rays 
and 
,![\[ \angle ABS=\angle ACR=180^\circ-\angle BGC.\]](http://latex.artofproblemsolving.com/0/a/d/0ad01e29194358ea0d3e10cb74b0769d1bd3a656.png)
Prove that 
.
and 
for which![\[ \left \lfloor \frac{a^2}{b} \right \rfloor + \left \lfloor \frac{b^2}{a} \right \rfloor = \left \lfloor \frac{a^2 + b^2}{ab} \right \rfloor + ab.\]](http://latex.artofproblemsolving.com/c/3/6/c3682397d21443cb720e56251c260b984392a84a.png)
,
,
,
.
meets 
at 
,
.
be the reflection of 
.
.
meets 
and 
at 
and 
,
meets 
.
,
,
form an arithmetic progression.
such that 
for all 
.
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
,![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
right and 
,
.
we take two points 
so that 
and 
.
and 
containing inside 
and 
,
and 
,
and 
,
and that 
,
we raise a segment 
perpendicular to the plane of the square. Show that the projections of 
,
,
and 
are coplanar if and only if ![$O\in [AC]\cup [BD]$](http://latex.artofproblemsolving.com/6/6/6/666df8f2afa87f15c79f9c96d9d1cea17a2a7af0.png)
.
.![\[
(2^x - 1)(5^x - 1) = y^n
\]](http://latex.artofproblemsolving.com/f/8/f/f8f013a96863bc5f39aa23edb98eed9aec54a17f.png)
have no positive integer solution 
.
Y by bharatputra, Royalreter1, hamup1, sicilianfan, MathematicsOfPi, Darn, jh235, yimingz89, abishek99, 15Pandabears, xantho, amburger66, AkshajK, TheMaskedMagician, MSTang, IsabeltheCat, spartan168, SHARKYBOY, Tan, DVA6102, pinetree1, flamefoxx99, thkim1011, droid347, hesa57, m-c-geometry, ahaanomegas, Binomial-theorem, jhoer, dantx5, DrMath, TheCrafter, rjiang16, bobthesmartypants, tigerat, csmath, 62861, librian2000, Eugenis, mathisawesome2169, inequality3206, PersonPsychopath, MathAwesome123, blep, joey8189681, thinkinavi, atmchallenge, Carrots101, yojan_sushi, Gmud, bearytasty, sampai7, yrnsmurf, shiningsunnyday, mathguy5041, ohmcfifth, hwl0304, elitechicken, epiclucario, Shaddoll, Wave-Particle, DeathLlama9, champion999, gradysocool, PiDude314, bestwillcui1, DivideBy0, m1234567, brianapa, niyu, oceanair, phi_ftw1618, SW1, hockey10, weilunsun28, mathfun5, BIGBUBBLE, Marzi, kk108, always_correct, Chessking345, Aspen, itised, ItzVineeth, jeffisepic, DaniyalQazi2, mathwiz0803, Ultroid999OCPN, jangkun217, Liopleurodon, AFK16, Heisenberg09, Mathisfun04, green_dog_7983, reckoner, ccx09, TheMagician, Math-Ninja, coolak, AIME12345, andyxpandy99, vsathiam, mathlogician, dchenmathcounts, mathleticguyyy, Vietjung, Williamgolly, Susanssluk, Mathscienceclass, csunllc, FIREDRAGONMATH16, smartninja2000, Imayormaynotknowcalculus, middletonkids, Smile2020, skyscraper, mathbw225, SlimTune, RedFireTruck, IceWolf10, superagh, v4913, centslordm, SuperJJ, fidgetboss_4000, sotpidot, Natsuki, vsamc, OliverA, suvamkonar, aie8920, fuzimiao2013, ihatemath123, GCA, hi.., kankan, LOGYC, RubixMaster21, jhu08, mahaler, megarnie, exp-ipi-1, math31415926535, eagles2018, YuyangZ, sugar_rush, son7, Bradygho, asimov, StopSine, Jndd, Lamboreghini, nikitadas, EpicBird08, s12d34, Adventure10, Paul10, ehuseyinyigit
I know many of us are looking for last-minute AIME practice around this time, since the AIME I is less than two weeks away. With that in mind, here is a collection of AIME level problems that can be used as practice. The problems are sorted into four categories (Algebra, Combinatorics, Geometry, Number Theory) with room for overlap, and there are ten problems in each category arranged (hopefully) from easiest to hardest. Depending on my strengths and weaknesses, some of the subjects have harder problems than others, and not all the problems are properly placed or of high quality, but hopefully this can be used to sharpen your skills before the actual AIME. Enjoy!
EDIT: Oh, and feel free to post solutions to any of the problems!
Edit (December 25, 2019): The solutions packet to this can be found on my website here.
EDIT: Oh, and feel free to post solutions to any of the problems!
Edit (December 25, 2019): The solutions packet to this can be found on my website here.
This post has been edited 2 times. Last edited by djmathman, Dec 26, 2019, 12:15 AM
.
but it asked for "the smallest integer greater than *insert expression here*", making it zero.
and 
.
as the number of ways to label the points if there are 
points, and all other conditions still hold. To find 
sections, and section 
has length 
.
to 
.
.
is prime, so it is very easy to apply the lemma.
.
we have![\[a^2+b^2=\frac{10^6}{a^2}+\frac{10^6}{b^2}=2008.\]](http://latex.artofproblemsolving.com/2/5/0/25010e1afa66cdf66683ca31949a661eea396517.png)
Then because 
![\[\frac{10^6}{2008-b^2}+\frac{10^6}{b^2}=2008.\]](http://latex.artofproblemsolving.com/7/8/e/78ee79c7ce3f77bc217fb5c8b7afc3b4268f1d0a.png)
![\[\rightarrow 10^6=b^2(2008-b^2)\]](http://latex.artofproblemsolving.com/d/c/9/dc9e016af42b0f2827c6d64c14f350a50ca88479.png)
Notice that 
as well, so the polynomial![\[x^2-2008x+10^6=0\]](http://latex.artofproblemsolving.com/b/9/d/b9d68deb110097d62bfde0dabe9bde67ea1d4864.png)
has roots 
.
and 
![\[a+b=\sqrt{4008},\ a-b=\sqrt{8}.\]](http://latex.artofproblemsolving.com/f/0/3/f0326d443e071af49d167c4446b6942c666f2d75.png)
which then means 
(because 
)
.
,
such that 
.![\[t^2=(\sqrt3+\sqrt2)^2=5+\sqrt{24}\implies t^2-5=\sqrt{24}.\]](http://latex.artofproblemsolving.com/3/8/a/38a23358440d1849b157e9e54ff4731d6d1efbc7.png)
Squaring again gives 
,
and 
.
and the requested answer is 
.![\[a_{n+1}-1=\sqrt{a_n^2-2a_n+3}=\sqrt{(a_n-1)^2+2}\implies (a_{n+1}-1)^2=(a_n-1)^2+2.\]](http://latex.artofproblemsolving.com/1/c/b/1cbfd071d2d6b5c642c70a969d55625e517c400c.png)
Let 
.
and 
,
for all positive integers 
,![\[(a_{513}-1)^2=1024\implies a_{513}-1=32\implies a_{513}=\boxed{033}.\]](http://latex.artofproblemsolving.com/9/e/3/9e34ac2e7225d292a372c153ed110f9bdb0e6269.png)
(Source: OMO Fall 2012, Ray Li)
gives ![\[x^2-\lfloor x^2\rfloor=(x-\lfloor x\rfloor)^2\implies \lfloor x^2\rfloor=2x\lfloor x\rfloor-\lfloor x\rfloor^2.\]](http://latex.artofproblemsolving.com/5/c/1/5c1cd57de32c54fc83336bdb211a7eddf67a9aca.png)
Let 
.
to be integral, so 
for some positive integer 
.![\[\{x^2\}=\left\{\left(k+\dfrac a{2k}\right)\right\}^2=\left\{k^2+a+\left(\dfrac a{2k}\right)^2\right\}=\left(\dfrac a{2k}\right)^2=\{x\}^2.\]](http://latex.artofproblemsolving.com/a/a/6/aa6df1d694faa33ad09ad1d10a448d70f7c98565.png)
Now it remains to find their sum. This is calculation; if we are careful, we should get that the sum equals ![\begin{align*}&99+\sum_{k=1}^{98}\left[k+\left(k+\dfrac1{2k}\right)+\left(k+\dfrac2{2k}\right)+\cdots+\left(k+\dfrac{2k-1}{2k}\right)\right]\\&=99+\sum_{k=1}^{98}\left[k(2k)+\dfrac1{2k}\sum_{a=0}^{2k-1}a\right]\\&=99+\sum_{k=1}^{98}\left[2k^2+\dfrac{(2k-1)(2k)}{2(2k)}\right]\\&=99+\sum_{k=1}^{98}\left[2k^2+k-\dfrac12\right]\\&=99+2\sum_{k=1}^{98}k^2+\sum_{k=1}^{98}k-\sum_{k=1}^{98}\dfrac12\\&=99+2\left[\dfrac{98(99)(197)}6\right]+\dfrac{98(99)}2-48\\&=641999.\end{align*}](http://latex.artofproblemsolving.com/c/d/a/cdab9cca97aa2896937b245a0872690016daa951.png)
Finding 
as opposed to 
.
,
.![\[(ac)^2+(ad)^2+(bc)^2+(bd)^2=2008^2.\]](http://latex.artofproblemsolving.com/c/2/7/c27ae0ba79a799eb81fc9ded4598b6a2c9c24822.png)
Next note that if 
are two equal numbers then 
.
and 
gives 
.![\[(ac)^2+(ad)^2+(bc)^2+(bd)^2=(ad)^2+2abcd+(bc)^2=(ad+bc)^2=2008^2,\]](http://latex.artofproblemsolving.com/0/7/7/07730baaf25be15b58c8cdd3069432cf61e5ebbb.png)
so 
,![\[a^2+b^2+c^2+d^2-2(ad+bc)=0\implies (a-d)^2+(b-c)^2=0,\]](http://latex.artofproblemsolving.com/0/2/c/02c6f38acb3cbd9127355344bf0d038b94630bfe.png)
implying that 
and 
.![\[(a+b)(c+d)=ac+bd+(ad+bc)=4008.\]](http://latex.artofproblemsolving.com/4/a/1/4a1fb02382e6d4ddfe2a4f45d3e064f6cff250bc.png)
From the equalities established above, we have 
,
.
?
or am I being dumb? Also how did you bound a? Thank you!
is an integer, so the fractional part of 
for some integer 
in the desired range. Then 
.
,
.
answer extraction as it should.
