<p>Here’s another problem:</p>
<p>Nine different, non-zero integers are placed into a 3x3 grid such that, for every row and column, the sum of the third number and the product of the first two numbers (reading left to right, top to bottom) is a constant. What is the minimum possible sum of the absolute values of the nine integers?</p>
<p>@ Dad of 3: don’t think an answer has been posted, but (m+n) choose m? (or m+n choose n, they’re equal) because you have a total of m+n moves to get there, of which you choose m of those to be going on the parallel streets and n to be on the perpendicular streets.</p>
<p>darn, can’t get rspence’s problem without guessing/checking…</p>
<p>here’s an easy one:</p>
<p>suppose x, y, and z are positive reals such that x + y + z = xyz. find the maximum value of (cyclic sum x/(sqrt(1+x^2))</p>
<p>I claim that the maximum value is 3 sqrt(3)/2.</p>
<p>Let x = tan A, y = tan B, z = tan C, where 0 ≤ A,B,C ≤ pi/2. Then x/(sqrt(1+x^2)) = (tan A)/(sqrt(1 + tan^2 A)) = (tan A)/(sec A) = sin A.</p>
<p>So now we are maximizing sin A + sin B + sin C given that tan A + tan B + tan C = (tan A)(tan B)(tan C). However there’s a theorem that says the “given” statement is true if and only if A,B,C are the angles of a triangle, so A+B+C = pi.</p>
<p>We will now prove that sin A + sin B + sin C is maximized when A = B = C = pi/3. Letting a,b,c be the sides of the triangle, a/(sin A) = b/(sin B) = c/(sin C) = 2R. If R is fixed, then</p>
<p>sin A + sin B + sin C = (a+b+c)/2R</p>
<p>Hence maximizing LHS is equivalent to maximizing the RHS, which I believe occurs when the three points are equally spaced around the circle (hence a = b = c, haven’t proven this yet).</p>
<p>If the above is true, the maximum value of the cyclic sum is 3 sin (pi/3), or 3 sqrt(3)/2.</p>
<p>Guys…these problems are more or less lv10+ problems now…</p>
<p>…and this thread begins to look like an AoPS marathon thread. lol</p>
<p>@Kyrix1 </p>
<p>My thoughts exactly. These are nice problems, but this is not the thread I intended. </p>
<p>That said, I like what I’m seeing, so I’m content to just sit back and see what unfolds. :)</p>
<p>Yeah, that last problem felt like a #1 or #2 on USAMO, and I had to look up the tan A + tan B + tan C = (tan A)(tan B)(tan C) theorem. Nice problem though.</p>
<p>Next up, an IMO #6 where Schur and Muirhead’s inequalities are completely useless. :D</p>
<p>From AoPS, the $ signs are LaTeX rendering:</p>
<p>Let $ ABC$ be an equilateral triangle of side length 1. Let $ D$ be the point such that $ C$ is the midpoint of $ BD$, and let $ I$ be the incenter of triangle $ ACD$. Let $ E$ be the point on line $ AB$ such that $ DE$ and $ BI$ are perpendicular. $ \Omega$ is the circumcircle of triangle $ DBE$ with center $ Z$. Let $ F$ be the point of intersection of $ \Omega$ with ray $ ZA$. Construct line $ l$, the reflection of the angle bisector of $ \angle BFE$ over line $ FZ$. Let $ l$ intersect $ EB$ at $ L$ and $ DE$ at $ L’$. Define the intersection of $ DL$ and $ BL’$ to be $ G$, and let $ EG$ and $ BD$ intersect at $ L’’$. Draw the line parallel to $ EZ$ passing through $ L’’$. Let the intersection of this line with $ \Omega$ closer to $ E$ be $ H$. Define point $ X$ such that $ LX$ is perpendicular to $ AE$, $ X$ is on the same side of $ AE$ as $ Z$, and quadrilaterals $ ALEX$ and $ ZHAI$ have the same area. Let $ J$ be the reflection of $ X$ across line $ FH$, and let $ J’$ be the image of $ J$ when inverted about $ \Omega$. If $ \omega$ is the circumcircle of triangle $ FJ’H$, construct $ T$ to be the point such that $ TF$ is the diameter of $ \omega$. Define $ W$ to be the point such that it lies on line $ TZ$ and it is the center of a circle internally tangent to both $ \Omega$ and $ \omega$. Let $ K$ be the Gergonne point of triangle $W TF$, and project $ K$ onto line $ BE$. Call this projection $ M$. Let $ N$ be the isotomic conjugate of $ M$ with respect to triangle $ IKE$. Define $ O<em>J$, $ O</em>K$, $ O<em>N$ to be circles whose centers are on the medians drawn from $ J, K, N$ in triangle $ JKN$. Suppose that these circles are all internally tangent to $ \omega$ and respectively tangent to $ KN, NJ, JK$. Let $ O</em>{M^G}$ be the circle internally tangent to the each of $ O<em>J, O</em>K, O<em>N$. Dilate the circumcircle of triangle $W TF$ so that it is coincident with circle $ O</em>{M^G}$. Let $ W’,T’,F’$ be image of vertices $ W,T,F$ under this dilation. Let $ M^G$ be the isogonal conjugate of the center of $ O<em>{M^G}$ in triangle $ W’T’F’$. Let line $ M^G O</em>N$ intersect $ ME$ at $ P$ and $ ED$ at $ L<em>E$ such that $ L</em>E$ and $ M^G$ are on opposite sides of $ O<em>N$. Construct regular heptadecagon $ P L</em>E A<em>S E</em>T E<em>L L</em>M E<em>Y O</em>U A<em>R E</em>N O<em>T D</em>R A<em>W I</em>N G<em>T H</em>I S$ such that $ S$ and $ Z$ are on the same side of line $ P L_E$. Let the radical axis of the circumcircles of triangles $ TEN$ and $ SIX$ intersect $ AM$ at $ R$. Let $ \mathcal{P}$ be the region formed by the convex hull of the points $ P,A,L,M,E,R$. Find the area of $ \mathcal{P}$</p>
<p>Apparently this was in the WOOT forum 2 years ago.
Answer = 0 (if I am to believe my sources)</p>
<p>Haha :D</p>
<p>I love the Alex Zhai reference.</p>
<p>Have you seen the mock SAT exam? With math problems ranging from SAT-level to IMO-level.</p>
<p>For those of you who haven’t checked it out:
<a href=“http://dl.dropbox.com/u/4538665/LaTeX/Other/Archive/MockSAT1Math/MockSAT1Math.pdf[/url]”>http://dl.dropbox.com/u/4538665/LaTeX/Other/Archive/MockSAT1Math/MockSAT1Math.pdf</a></p>
<p>Yep. For the math section, I’ve solved 21-31 except for 27 (30 was on last year’s AMC). 32 and on I’m like, this is too hard. Although, 34 looks pretty easy. Lol, pretty sad I actually tried solving the problems.</p>
<p>And then the essay prompt. Although one could just cite the paper by Andrew Wiles.</p>
<p>wait Kyrix1 you were in WOOT two years ago? hey so was i! i also saw evan’s math sat 
so this thread turns into aops people presenting especially beautiful problems…</p>
<p>edit: @rspence you were right, i used jensen to finish it up lol</p>
<p>Haha 
</p>
<p>Btw I also took WOOT two years ago (2010-2011).</p>
<p>@funsummer
Nah, I wan’t in the WOOT 2 years ago (starting it this year) :\ although I know someone who was, and he shares this stuff with me; he’s going for USAMO winner this year. :D</p>
<p>OK BACK TO EASY PROBLEMS:
Let AB = 14 be a diameter of circle O, and let C be a point on the circle such that angle(BAC) = 60 degrees. Find the length of BC.</p>
<p>That’s like a level 3 problem.</p>
<p>Level 5-6 problem:
A point P is randomly chosen in square ABCD. What is the probability that angle APB < 90?</p>
<p>Yours uses the exact same concept as mine -.-</p>
<p>Yeah, I intentionally made it similar (involving circles) but slightly harder. Although they’re both easy problems…</p>
<p>Another easy one:</p>
<p>What is the sum of all positive integers n such that (n^2 + 2012)/(n - 2) is an integer?</p>
<p>These last couple of problems are more of what I had in mind for this thread. Let’s see if we can keep them at this level, and post the harder ones in my Challenge Thread.</p>
<p>@rspence: I’m not very sure, but is the answer to that 1014?
Probability: 1- 0.25 pi ?
Length of BC: 7 sqrt 3 ?</p>
<p>Btw, how I quote people on this forum? (pardon me if the question sounds inane. :P)</p>
<p>@yellowcat Rewrite as (n^2 - 4)/(n-2) + 2016/(n-2). Since (n^2 - 4)/(n-2) is an integer, so is 2016/(n-2) and it follows that n-2 is a factor of 2016.</p>
<p>sigma(2016) = 6552. However we have to add 2 to each factor of 16 to find n. There are 36 factors of 16 (this is done by finding prime factorization), and 6552 + 2(36) = 6624. That’s what I got…</p>