It's been a while

by djmathman, Apr 1, 2025, 10:05 PM

and I don't have to much to say, except this.

djmathman, 17 October 2020 wrote:

My hope is that, now that I'm in my second year of graduate school, I can shift toward completing my doctorate degree (or even figuring out what area of math I want to study in the first place) and spend less time on competition math as a whole.

https://static.tvtropes.org/pmwiki/pub/images/chrome_2018_01_16_21_51_41.png

This post has been edited 2 times. Last edited by djmathman, Apr 1, 2025, 10:21 PM

2024 AMC Proposals

by djmathman, Nov 13, 2024, 11:17 PM

Four proposals this time. One great, three OK.

12A 12/10A 19. The first three terms of a geometric sequence are the integers $a,\,720,$ and $b,$ where $a<720<b.$ What is the sum of the digits of the least possible value of $b?$

Comments

10A 22. Let $\mathcal K$ be the kite formed by joining two right triangles with legs $1$ and $\sqrt3$ along a common hypotenuse. Eight copies of $\mathcal K$ are used to form the polygon shown below. What is the area of triangle $\triangle ABC$ ?

https://cdn.artofproblemsolving.com/attachments/1/3/03abbd4df2932f4a1d16a34c2b9e15b683dedb.png

Comments

10A 25/12A 22. The figure below shows a dotted grid $8$ cells wide and $3$ cells tall consisting of $1''\times1''$ squares. Carl places $1$ -inch toothpicks along some of the sides of the squares to create a closed loop that does not intersect itself. The numbers in the cells indicate the number of sides of that square that are to be covered by toothpicks, and any number of toothpicks are allowed if no number is written. In how many ways can Carl place the toothpicks? $[asy] size(6cm); for (int i=0; i<9; ++i) { draw((i,0)--(i,3),dotted); } for (int i=0; i<4; ++i){ draw((0,i)--(8,i),dotted); } for (int i=0; i<8; ++i) { for (int j=0; j<3; ++j) { if (j==1) { label("1",(i+0.5,1.5)); }}} [/asy]$ $\textbf{(A) }130\qquad\textbf{(B) }144\qquad\textbf{(C) }146\qquad\textbf{(D) }162\qquad\textbf{(E) }196$

Comments

12B 19. Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$ , where $0 < \theta < 60^{\circ}$ , to form $\triangle DEF$ . See the figure. The area of hexagon $ADBECF$ is $91\sqrt{3}$ . What is $\tan\theta$ ?
$[asy] defaultpen(fontsize(13)); size(170); pair O=(0,0),A=dir(225),B=dir(-15),C=dir(105),D=rotate(38.21,O)*A,E=rotate(38.21,O)*B,F=rotate(38.21,O)*C; draw(A--B--C--A,gray+0.4);draw(D--E--F--D,gray+0.4); draw(A--D--B--E--C--F--A,black+0.9); dot(O); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); [/asy]$
Comments

This post has been edited 2 times. Last edited by djmathman, Nov 13, 2024, 11:19 PM

AMC

Proof of AM-GM by Induction

by djmathman, Aug 7, 2024, 7:57 PM

and not Cauchy Induction. This proof can be found in Chapter 27 of the kind of insane book Classical and New Inequalities in Analysis.

Let $f_n(a)$ be the maximum value of $\textstyle\prod_{j=1}^n x_j$ subject to the conditions $x_j \geq 0$ and $\textstyle\sum_{j=1}^n x_j = a$ . We will crucially use the fact that $f_n$ is homogeneous, i.e. $f_n(a) = a^n f_n(1)$ .

Assume by scaling that $x_1 + \cdots + x_n = 1$ . Fix $x_n = t\in[0,1]$ . Then $\textstyle\sum_{j=1}^{n-1} x_j = 1 - t$ , and so the product $\textstyle\prod_{j=1}^n x_j$ is at least
$t f_{n-1}(1 - t) = t(1 - t)^{n-1} f_{n-1}(1).$ The minimum value of $t(1-t)^{n-1}$ in the interval $t\in[0,1]$ is $\tfrac{(n-1)^{n-1}}{n^n}$ , achieved when $t = \tfrac 1n$ . (This is easiest to prove by differentiating the analogous expression $(1-t)t^{n-1} = t^{n-1} - t^n$ .) Therefore
$f_n(1) = \max_{t\in[0,1]}t(1-t)^{n-1}f_{n-1}(1) = \frac{(n-1)^{n-1}}{n^n}f_{n-1}(1).$ A quick induction argument implies $f_n(1) = \tfrac{1}{n^n}$ , which is equivalent to AM-GM.

This post has been edited 1 time. Last edited by djmathman, Aug 7, 2024, 7:57 PM

It's been a while since I've posted math on here

by djmathman, Jul 19, 2024, 1:21 AM

so have a random collection of problems I've solved over the past $N$ months, where $N\approx 12$ .

CMIMC 2024 Algebra 9 (Robert Trosten). Let $\mathbb Q_{\geq 0}$ be the non-negative rational numbers, $f: \mathbb Q_{\geq 0} \to \mathbb Q_{\geq 0}$ such that $f(z+1) = f(z)+1$ , $f(1/z) = f(z)$ for $z\neq 0$ , and $f(0) = 0.$ Define a sequence $P_n$ of non-negative integers recursively via $P_0 = 0,\quad P_1 = 1,\quad P_n = 2 P_{n-1}+P_{n-2}$ for every $n \geq 2$ . Find $f\left(\frac{P_{20}}{P_{24}}\right).$

Solution

CMIMC 2024 Algebra 10 (Connor Gordon). There exists a unique pair of polynomials $(P(x),Q(x))$ such that
$\begin{align*} P(Q(x))&= P(x)(x^2-6x+7) \\ Q(P(x))&= Q(x)(x^2-3x-2) \end{align*}$ Compute $P(10)+Q(-10)$ .

Solution

Romania District Olympiad 2024 Grade 12 Problem 4. Let $f:[0,\infty)\to\mathbb{R}$ be a differentiable function, with a continous derivative. Given that $f(0)=0$ and $0\leqslant f'(x)\leqslant 1$ for every $x>0$ prove that $\frac{1}{n+1}\int_0^af(t)^{2n+1}\mathrm{d}t\leqslant\left(\int_0^af(t)^n\mathrm{d}t\right)^2,$ for any positive integer $n{}$ and real number $a>0.$

Solution

OTIS Mock AIME Problem 15 (Wilbert Chu). A parabola in the Cartesian plane is tangent to the $x$ -axis at $(1,0)$ and to the $y$ -axis at $(0,3)$ . Find the sum of the coordinates of the vertex of the parabola.

Solution

OMMC PoTM March 2022 (Evan Chang). Define acute triangle $ABC$ with circumcircle $\omega.$ Let $Q$ be the midpoint of minor arc $BC$ in $\omega$ and let $Q'$ be the reflection of $Q$ over $BC.$ If the circle with diameter $BC$ is tangent to the external angle bisector of $\angle BAC$ at $P,$ show $\angle BPQ' = \angle CPA.$

Solution

BAMO 2010 Problem 4. Acute triangle $ABC$ has $\angle BAC < 45^\circ$ . Point $D$ lies in the interior of triangle $ABC$ so that $BD = CD$ and $\angle BDC = 4 \angle BAC$ . Point $E$ is the reflection of $C$ across line $AB$ , and point $F$ is the reflection of $B$ across line $AC$ . Prove that lines $AD$ and $EF$ are perpendicular.

Solution

Naoki Sato. In triangle $ABC$ , sides $AB$ and $AC$ are extended to $D$ and $E$ , respectively, so that $DE$ is parallel to $BC$ and tangent to the $A$ -excircle. Construct the circle that passes through $D$ and $E$ , and is tangent to the $A$ -excircle. Prove that this circle is also tangent to the incircle of triangle $ABC$ .
$[asy] real area(pair A, pair B, pair C) { return(abs((xpart(A)*ypart(B) + xpart(B)*ypart(C) + xpart(C)*ypart(A) - ypart(A)*xpart(B) - ypart(B)*xpart(C) - ypart(C)*xpart(A)))/2); }; pair excentre(pair A, pair B, pair C) { return((-abs(B - C)*A + abs(C - A)*B + abs(A - B)*C)/(-abs(B - C) + abs(C - A) + abs(A - B))); }; real exradius(pair A, pair B, pair C) { return(2*area(A,B,C)/(-abs(B - C) + abs(C - A) + abs(A - B))); }; path excircle(pair A, pair B, pair C) { return(Circle(excentre(A,B,C),exradius(A,B,C))); }; unitsize (0.28 cm); pair A, B, C, D, E, P, T, X; A = (0,0); B = (-1,-4); C = (5,-4); D = interp(A,B,exradius(A,B,C)/inradius(A,B,C)); E = interp(A,C,exradius(A,B,C)/inradius(A,B,C)); X = (-9.19135,-14.6049); T = (incenter(A,D,E) + reflect(D,E)*(incenter(A,D,E)))/2; P = reflect(X,incenter(A,D,E))*(T); draw(incircle(A,B,C),red); draw(excircle(A,B,C),red); draw(circumcircle(D,E,P),blue); draw(A--D--E--cycle); draw(B--C); label("$A$", A, N); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, SE); [/asy]$
Solution

This post has been edited 2 times. Last edited by djmathman, Jul 19, 2024, 6:19 PM

I wrote a CMIMC problem this year

by djmathman, Apr 19, 2024, 12:49 PM

Combo/CS 1. For each positive integer $n$ (written with no leading zeros), let $t(n)$ equal the number formed by reversing the digits of $n$ . For example, $t(461) = 164$ and $t(560) = 65$ . For how many three-digit positive integers $m$ is $m + t(t(m))$ odd?

I wonder how many people got the reference/"point" of the problem during the test. (In particular, that second example uses a particularly specific number....)

This post has been edited 1 time. Last edited by djmathman, Apr 19, 2024, 12:51 PM

Geometric Proof of Heron's Formula

by djmathman, Feb 23, 2024, 1:25 AM

Here is a proof of Heron's Formula that almost exclusively makes use of similar triangles. This proof was recorded in [1] by E. E. Whitford and is based on a different (uncited) solution by Legendre.

We'll need a bit of setup. In triangle $ABC$ , assume without loss that $AB \leq AC$ . Choose points $D$ and $E$ on $\overline{AC}$ and $\overrightarrow{AB}$ , respectively, such that $AD = AB$ and $AC=AE$ . The midpoints $M$ and $N$ of $\overline{DB}$ and $\overline{EC}$ , respectively, are collinear with $A$ , and in fact $AMN$ is perpendicular to both segments.

Let $O$ be the midpoint of $\overline{DC}$ , and let $MO$ meet $CE$ at a point $G$ . Because triangle $\triangle MNG$ is right, the circle with center $O$ and radius $MO=GO$ also passes through $N$ . Finally, suppose this circle intersects $AC$ at two points $J$ and $K$ with $AJ < AK$ .

$[asy] size(200); defaultpen(linewidth(0.6)+fontsize(11)); pair A = (0,6), B = (2,0), C = (-3,-3), M = (0,0), N = (0,-3), D = (-2,0), E = (3,-3); pair O = (C+D)/2, G = M+C-B; real r = abs(M-O); path circ = circle(O,r); pair J = O + r * dir(aTan(3)), K = O - r * dir(aTan(3)); draw(A--B--C--A,rgb(0.1,0.6,0.1)); draw(C--E--B--D,rgb(0.1,0.1,0.7)); draw(M--G--C--K,rgb(0.7,0.1,0.7)); draw(A--N,rgb(0.9,0.3,0.1)+dashed); draw(circ, rgb(0.9,0.1,0.1)); dot("$A$",A,dir(90),linewidth(3.3)); dot("$B$",B,dir(0),linewidth(3.3)); dot("$C$",C,SE,linewidth(3.3)); dot("$D$",D,W,linewidth(3.3)); dot("$E$",E,SE,linewidth(3.3)); dot("$M$",M,NE,linewidth(3.3)); dot("$N$",N,SE,linewidth(3.3)); dot("$O$",O,NW,linewidth(3.3)); dot("$G$",G,S,linewidth(3.3)); dot("$J$",J,NW,linewidth(3.3)); dot("$K$",K,SW,linewidth(3.3)); [/asy]$
This setup may seem contrived, but the following lemma will elucidate the purpose of circle $O$ .

Lemma.Using standard triangle notation ( $BC = a$ , etc.), we have
$\begin{array}{ll} AK = s, & AJ = s - a,\\ CK = s - b, & CJ = s - c. \end{array}$ Proof. Observe that $AD = c$ , $JO = KO = MO = \tfrac12a$ , and $AO = \tfrac12(AD+AC) = \tfrac12(b+c)$ . From this, we may read off all four equalities:
$\begin{align*} AK &= AO + KO = \tfrac12(b+c) + \tfrac12a = s,\\ AJ &= AK - KJ = s - 2KO = s - a,\\ CK &= DJ = AJ - AD = s - b,\\ CJ &= AC - AJ = s - c. \end{align*}$ This completes the proof. $\square$

Returning to the main proof, remark that
$[ABC] = [ACE] - [BCE] = CN\cdot AN - CN\cdot MN = CN\cdot AM.$ Furthermore, triangles $ADM$ and $ACN$ are similar, so $CN\cdot AM = AN\cdot DM = AN\cdot GC$ . It follows from these manipulations and Power of a Point that
$\begin{align*} [ABC] &= \sqrt{(CN\cdot AM)^2} \\ &= \sqrt{(CN\cdot AM)\cdot (AN\cdot GC)}\\ &= \sqrt{(CN\cdot CG)\cdot(AM\cdot AN)}\\ &= \sqrt{(CJ\cdot CK)\cdot (AJ\cdot AK)}\\ &= \sqrt{s(s-a)(s-b)(s-c)}.\qquad\blacksquare \end{align*}$
References.
[1] (1914) Problems and Solutions, The American Mathematical Monthly, 21:7, 226-239, DOI: 10.1080/00029890.1914.11998045

This post has been edited 2 times. Last edited by djmathman, Feb 24, 2024, 8:22 PM

2023 AMC Proposals

by djmathman, Nov 15, 2023, 6:52 PM

I had quite a few this year! Most of them were on the easy end, though. That's part of the natural progression of things, I guess; I've been less interested in making hard problems and more focused on the easier end of the exam. Though this year it looks like the problems on the hard end were not great; I have some thoughts about a few of them....

12A 6 (March 2023). Points $A$ and $B$ lie on the graph of $y=\log_{2}x$ . The midpoint of $\overline{AB}$ is $(6, 2)$ . What is the positive difference between the $x$ -coordinates of $A$ and $B$ ?

10A 12 (March 2023). How many three-digit positive integers $N$ satisfy the following properties?

The number $N$ is divisible by $7$ .
The number formed by reversing the digits of $N$ is divisible by $5$ .

12A 16 (March 2023). Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$ . The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?

10B 1/12B 1 (March 2023). A mom is pouring orange juice for her 4 kids into 4 identical glasses. She fills the first 3 full, but only fills one third of the glass for the last one. How much does she need to pour from the 3 full glasses to fill all of the glasses to an equal amount?

10B 5 (March 2023?). Maddy and Lara see a list of numbers written on a blackboard. Maddy adds $3$ to each number in the list and finds that the sum of her new numbers is $45$ . Lara multiplies each number in the list by $3$ and finds that the sum of her new numbers is also $45$ . How many numbers are written on the blackboard?

10B 7 (March 2023). Square $ABCD$ is rotated $20^\circ$ clockwise about its center to obtain square $EFGH$ , as shown below. What is the degree measure of $\angle EAB$ ?
$[asy] size(170); defaultpen(linewidth(0.6)); real r = 25; draw(dir(135)--dir(45)--dir(315)--dir(225)--cycle); draw(dir(135-r)--dir(45-r)--dir(315-r)--dir(225-r)--cycle); label("$A$",dir(135),NW); label("$B$",dir(45),NE); label("$C$",dir(315),SE); label("$D$",dir(225),SW); label("$E$",dir(135-r),N); label("$F$",dir(45-r),E); label("$G$",dir(315-r),S); label("$H$",dir(225-r),W); [/asy]$

12B 12 (December 2020). For complex numbers $w=a+bi$ and $z=c+di$ , define the binary operation $\otimes$ by $w\otimes z=ac+bd\,i.$ Suppose $z$ is a complex number such that $z\otimes z=z^{2}+40$ . What is $|z|$ ?

You'll notice a lot of March 2023 problems here. Some of you may also remember that I wrote the following last year:

Quote:

In any case, that'll be it for a while. As I mentioned before, I didn't submit anything for the 2023-2024 series of contests, so you probably won't see any dj problems there. I probably won't be completely dead, but I certainly don't feel the same passion and fire for problem construction as I used to. (Right now, I'm more focused on improving my mental well-being anyway.)

It turns out that right before the AMC constructor meetings last year I had a sudden interest in constructing some problems for the test. The AMC heads requested some complex numbers and logarithm problems, so I focused my efforts on those, with a few extra problems thrown in there for good measure. This was the result of about two weeks' worth of thought, putting research to the side for a bit. There's nothing super flashy here, but I'm decently happy with this set. Favorite is almost certainly 12A #16, though 10B #7 is up there as one of my favorites, too.

This post has been edited 1 time. Last edited by djmathman, Nov 15, 2023, 6:55 PM

AMC

A Generalization of the Bernoulli Inequality

by djmathman, Oct 7, 2023, 6:44 PM

Here's a nice generalization of Bernoulli's Inequality I found the other day. This result was originally recorded in [1] (possibly earlier, I honestly can't tell!) and is, in my opinion, less known than it should be.

Recall the statement of Bernoulli's Inequality: $(1+x)^r \geq 1 + rx$ whenever $x \geq -1$ and $r \geq 1$ (with the reverse inequality occuring when $0 < r < 1$ ). In certain special cases, this is easy to see; for example, when $n = 3$ and $x \geq 0$ ,
$(1+x)^3 = 1 + 3x + 3x^2 + x^3 \geq 1 + 3x.$ However, such a naive process does not necessarily work when $r$ is not an integer. For example, consider the case $r = \tfrac12$ , which has Taylor expansion
$\sqrt{1+x} = 1 + \frac x2 - \frac18x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \cdots.$ Here the Taylor expansion consists of both positive and negative coefficients, so more subtle analysis is needed to make the above approach work. Nevertheless, Bernoulli's Inequality does say that $\sqrt{1+x} \leq 1 + \tfrac x2$ for all $x\geq -1$ .

At this point, we raise a natural question: does a "quadratic" version of Bernoulli's Inequality
$(1+x)^r \geq 1 + rx + \frac{r(r-1)}2x^2\qquad(1)$ hold in general? What about higher approximations? It turns out the general criterion is surprisingly simple.

Theorem ([1]). Let $\alpha$ be an arbitrary real number, and consider a hypothetical inequality of the form
$(1+x)^\alpha\ \square \ 1 + \alpha x + \binom\alpha2 x^2 + \cdots + \binom\alpha k x^k.\qquad(2)$ (Here $\textstyle\binom\alpha k = \tfrac{\alpha(\alpha - 1)\cdots (\alpha - k + 1)}{k!}$ , which allows for $\alpha$ to be a non-integer.) Assume $x \geq -1$ . Then the following hold.

If the next omitted term $\textstyle\binom\alpha{k+1}x^{k+1}$ of $(2)$ is positive, the LHS is greater than the RHS;
If the next omitted term is zero, the LHS and RHS are equal;
If the next omitted term is negative, the LHS is less than the RHS.

This means e.g. that, in the case $x \geq 0$ , $(1)$ holds iff $r \geq 2$ .

Proof. The reference [1] uses a proof by induction, but we'll instead go for a proof via Taylor's Theorem with Remainder. For ease of typesetting, let $P(x)$ be the right hand side of $(2)$ . Recall that $P(x)$ is the $k^{\text{th}}$ order Taylor polynomial for the function $f(x) = (1+x)^\alpha$ . It follows that
$\begin{align*} (1+x)^\alpha &= P(x) + \frac{f^{(k+1)}(\xi)}{(k+1)!}x^{k+1} \\ &= P(x) + \binom{\alpha}{k+1}(\xi + 1)^{\alpha-k-1}x^{k+1} \end{align*}$ for some real number $\xi$ between $0$ and $x$ . By assumption, $\xi \geq -1$ , so $(\xi + 1)^{\alpha - k - 1}$ is nonnegative. That is,
$\begin{align*} (1+x)^\alpha > P(x)\quad&\text{ if and only if}\quad \textstyle\binom{\alpha}{k+1}(\xi + 1)^{\alpha-k-1}x^{k+1} > 0\\ &\text{ if and only if}\quad \textstyle\binom\alpha{k+1}x^{k+1} > 0. \end{align*}$ This proves part 1 of the original theorem, and parts 2 and 3 follow by analogous reasoning. $\blacksquare$

References:

[1] Gerber, L. An Extension of Bernoulli's inequality. American Mathematical Monthly 75, 875 - 876 (1968).

This post has been edited 5 times. Last edited by djmathman, Sep 16, 2024, 4:55 PM

algebra

Zeros of Rational Function

by djmathman, Apr 9, 2023, 6:50 PM

Hey, more math.

"Complex Polynomials" Problem 2.1.7.5. Let $c_k > 0$ , $|z_k| \leq 1$ ( $1\leq k\leq n$ ) and set
$R(z) = \sum_{k=1}^n\frac{c_k}{(z-z_k)^m},$ where $m\geq 1$ is a natural number. Show that $R(z)$ has all its finite zeros in $\{|z| \leq \tfrac{1}{2^{1/m} - 1}\}$ .

Solution
Remark

This post has been edited 4 times. Last edited by djmathman, Apr 9, 2023, 6:56 PM

An Inequality with Harmonic Numbers

by djmathman, Mar 2, 2023, 9:26 PM

Whoa, actual math on this blog? Preposterous.

AMM E 819 (1948), H.F. Sandham. Let $\mathcal H_n$ denote the $n^{\text{th}}$ harmonic number. Prove that
$\gamma < \mathcal H_p + \mathcal H_q - \mathcal H_{pq} \leq 1,$ where $\gamma$ is Euler's constant.

Solution

This post has been edited 2 times. Last edited by djmathman, Mar 2, 2023, 9:27 PM

Submit

dj so orz
by Yiyj1, Mar 29, 2025, 1:42 AM
legendary problem writer
by Clew28, Jul 29, 2024, 7:20 PM
orz $\,$
by balllightning37, Jul 26, 2024, 1:05 AM
hi dj $\text{[math]}$ $\text{[math]}$
by OronSH, Jul 23, 2024, 2:14 AM
i wanna submit my own problems lol
by ethanzhang1001, Jul 20, 2024, 9:54 PM
hi dj, may i have the role of contributer?
by lpieleanu, Feb 23, 2024, 1:31 AM
This was helpful!
by YIYI-JP, Nov 23, 2023, 12:42 PM
waiting for a recap of your amc proposals for this year
by ihatemath123, Feb 17, 2023, 3:18 PM
also happy late bday man! i missed it by 2 days but hope you are enjoyed it
by ab456, Dec 30, 2022, 10:58 AM
Contrib?
by MC413551, Nov 20, 2022, 10:48 PM
tfw kakuro appears on amc
by bissue, Aug 18, 2022, 4:32 PM
Hi dj
by 799786, Aug 10, 2022, 1:44 AM
Roses are red,
Wolfram is banned,
The best problem writer is
Djmathman
by ihatemath123, Aug 6, 2022, 12:19 AM
hello
by aidan0626, Jul 26, 2022, 5:49 PM
Do you have a link to your main blog that you started after graduating from high school, I couldn't find it. @dj I met you IRL at Awesome Math summer Program several years ago.
by First, Mar 1, 2022, 5:18 PM
how did you do that cursor
by mathgenius64, Sep 1, 2011, 10:14 PM
Finally posted that problem! And hooray for pencil cursor!
by djmathman, Aug 24, 2011, 6:49 PM
Geo 2 is correct though
by sjaelee, Aug 24, 2011, 12:21 AM
Hello...Lol yesterday sjaelee posted Geomretry #2. Isn't it supposed to be Geometry #2? Lol.
by mcqueen, Aug 23, 2011, 11:57 PM
Delete geo #4! Plese!
by sjaelee, Aug 22, 2011, 11:41 PM
Hrm, there's this problem that I've been too lazy to post. I think I should soon.
by djmathman, Aug 22, 2011, 7:02 PM
Why hello...funny I'm saying this when I know you'll be unresponsive because you're V/LA for a week.
by mcqueen, Aug 14, 2011, 10:47 AM
But that's because those problems are the easiest to find.
by djmathman, Aug 12, 2011, 3:49 AM
Yea, I kind of keep forgetting to change that.
by djmathman, Aug 11, 2011, 11:27 AM
I noticed that only algebra and number theory come up (trig was basically algebra)
by sjaelee, Aug 11, 2011, 11:22 AM
There you go.
by djmathman, Aug 7, 2011, 2:07 PM
*cough* contrib
by Mrdavid445, Aug 7, 2011, 12:40 PM
Ah, i'm sure you will use this blog well.
by tc1729, Aug 7, 2011, 2:10 AM
First shout!
by djmathman, Aug 5, 2011, 6:58 PM

363 shouts

Blog of the (Former) Rising Olympian