Wayne Burrows

@JeffOstler1 @DanBaril33 Precisely this. I can't tell where the ball was and what his backswing path was before and after. Far from clear anything was improved.

Nobody is arguing that he stomped the grass. Thats not a penalty unless it improves his lie, his stance or his swing path. Bryson claims he was well behind the ball when he was stomping on the grass and it didn’t affect any of those. I’ve seen no video evidence to contradict him. Where was the ball when he did this and what was the distance between him and the ball? This is not “cut and dry.”

Anyone who thinks Bryson DeChambeau didn’t do what he did on purpose, or thinks he did not deserve the two stroke penalty or worse, does not know the game of golf. As far as I am concerned he is Patrick Reed 2.0.

@cicicrystal_ @NoContextHumans His arm was outstretched and the referee walked into him. Doesn't look like he grabbed at the ref to me.

@NoContextHumans The referee said, “Don’t touch me. I’m sensitive today. I’m in a bad mood”

The harshest red card you’ll ever see

A fixed circle and a line. Draw circles tangent to each as shown. What curve do the centers of those circles outline?

@jamestanton The blue outline is an equilateral triangle. The red triangle is approximately equilateral with integer grid point vertices. The black triangle is equilateral triangle shifted by half the distance to the apex of the grid triangle.

@jamestanton Overlaid blue equilateral triangle (0,0), (152,0), (76,76*sqrt(3)) and red outline (0,0),(152,0),(76,132). Then an enlargement at the apex to show the error.

For any square lattice of congruent dots centered on points with integer coordinates, it is possible to draw an equilateral triangle, horizontal base, with each corner landing within/on a dot. (How?) If each dot has radius r, what is the smallest side length of such a triangle?

#math problem 12-07-2026 Series equation🏁🟰

@jamestanton whoops bad english :D

@TannerCarawan @AnalysisFact Whoops I have an n-1 instead of n-2 in the last line

Let p(x) be a polynomial with only real roots. Then p'(x)² >= p(x) p''(x) for all x.

@Mike_Blackley Is this sex or the hypermodern gender. I am happy to disclose my sex but often refuse to play the gender game and please ignorance

A majority of Scotland’s MSPs have refused to say whether they are male or female. The Scottish Parliament now lists the sex of 73 MSPs as “undisclosed”. dailymail.com/news/article-1…

@jamestanton The number of circles is 2/3 the number of stars. That is a statement about ratio.

A Pedagogical Question: The ratio of circles to stars in this picture is "2 to 3." People say that this ratio can also be expressed 2:3 or as 2/3. Umm. Where exactly is the fraction 2/3 in this picture? In what sense, exactly, is a fraction a ratio?

@mrsrosieb @AuntyHeihei Seems racist to criticise him based on his ethnicity

@AuntyHeihei The first thing to do is to stop calling Seymour maori. He’s basically a white man with invisible drops of Maori blood.

How very raaaaacist of ACT to have two Māori leaders 🤣🤣.. But what it does highlight is that Maori are as capable of being as successful as anyone else... And that those clowns at TPM don't speak for all Māori...just like I said nearly 3 years ago..... Now let's drop the race based policies and bullshit and get the mahi done because power and money and the love of it will always be the problem, not race plainsight.nz/no-mandate-to-…

@MathMath901 Not convinced about my previous answers and so deleted them S=x+xS => S=x/(1-x) is not necessarily valid. S=x/(1-x)<-1 for x>1 but S=x+x(x+x(x+...))>1 for x>1 - actually it is infinite for all x>1 Say x=2 S=2/(1-2)=-2 S=2+2(2+2(2+...=2+4+8+... not less than -1

#math problem 24-06-2026 Algebra: Fixed Point problem.🔳

@jamestanton No

Let's say two natural numbers are "binary disjoint" if their binary representations don't share the digit 1 in the same column. (E.g. twenty=10100 and nine=1001 are binary disjoint.) In this picture, cell (a,b) is red if a and b are binary disjoint. Is this picture surprising?

@Suitandtie9999 Dishonest public transport policy. Dishonest doctor visit policy. Not sure is there a pattern here?

Labour’s election policies have quickly descended into farce. Chris Hipkins has now admitted that the party’s flagship promise of “three free doctors’ visits a year” might not involve seeing a doctor at all. According to Labour’s own policy, these so-called “free doctors’ visits” can be used with nurse practitioners instead. The party openly admits this, yet it cannot even say how many of those visits will actually be with a real doctor. Meanwhile, Labour is marketing the policy as “free doctors’ visits” to win votes, only to quietly concede it’s just another primary care appointment. What doesn’t change, however, is that taxpayers will still be forced to pay for it through the party’s proposed capital gains tax.

@jamestanton I think the primes are the standard counter example to this. sum (1/p) diverges but asymptotically the number of primes is n/log(n) so (n/log(n))/n->1/log(n)->0

A={a,b,c,d,...} is a subset of the counting numbers. E(n)=the number of elements of A that are less than or equal to n. If E(n)/n --> 0 as n grows, does that mean that the infinite sum 1/a + 1/b + 1/c + 1/d + ... converges? That is, does "sparsity imply convergence"?

@jamestanton Follows from 1-1/2+1/3-1/4+1/5-1/6+... = ln(2)

Prove, that on average, a number has ln(2)~0.69 more odd factors than even factors! (To be precise: Let F(n) = |odd factors of n| - |even factors of n|. Compute limit (F(1)+F(2)+...+F(n))/n as n grows.)

@jamestanton Follows from 1/1+1/3+1/6+...+2/n(n+1)+...=2 1/1+(2/2-2/3)+(2/3-2/4)+(2/4-2/5)+...+(2/n-2/(n+1)+(2/(n+1)-2/(n+2)+... =1/1+2/2+(-2/3+2/3)+)+(-2/4+2/4)+... =1+1+0+0+0+... =2

Prove that, on average, each number has 2 triangular numbers as factors. [To be precise, let F(n) = number of triangular factors of n and examine (F(1)+F(2)+...+F(n))/n as n grows.]

@EshaAA33 Long shot: they don't run businesses that earn enough.

Why are women missing from the table and will this ever change?😔

@Suitandtie9999 Targets 1.3million users. Average savings of $25 per week. 1.3million x 25 x 52 = $1.69 billion. Are Labour and Tangi Utikere mathematically innumerate, or are the numbers in their policy document dishonest?

Here we go again. Appearing on Q+A, Labour’s Transport spokesperson Tangi Utikere told Jack Tame that their public transport fare cap policy will be funded by taking money from road maintenance — including the repair of potholes. Under the last Labour government, potholes became such a serious problem for road users that National campaigned heavily on the issue and since the coalition government took office, there has been clear and significant progress. Labour taking money from road maintenance will undo all that work.