Recent content by no_arg

Hi all I am after as many 2020 trial papers (actual papers sat by students in 2020) with schemes/solutions that I can get a hold of, 2U,3U and 4U. If I get enough I will set up a grid where you pick off particular topics in prep for the HSC and to assist others in the future. Are they already...

Yes, these sorts of questions often appear in trial papers as if they were trivial and unique, when in fact they are generally only solvable by exhaustion with multiple solutions. Could there ever be more than 8 solutions? See Singmaster Conjecture.

Find all n and k such that (n choose k) is the sum of (13 choose 6) and (13 choose 5).

The use of congruence makes it less direct.

Simply rotate the entire diagram about P anticlockwise by sixty degrees and voila!

Both proofs a little clumsy

Given similarity, the solution cannot depend on the side length of the original triangle. Answer is disturbingly simple and the proof is stunning.

No X can be any interior point. PX QX and RX need to be lifted out of the diagram to form another triangle.

Suppose that PQR is an equilateral triangle with an interior point X. Let angle PXR = s and angle QXR = t. Find (in terms of s and t) the three angles of any triangle with side lengths equal to PX, QX and RX.

I think you can have x pies but you can't have pi x's!

Let x>0.. x^2=x+x+\dots+x \qquad where the sum on the right has x terms. For example if x=4 we have 4^2=4+4+4+4. Differentiating \qquad x^2=x+x+\dots+x \qquad we have 2x=1+1+\dots 1 \qquad where the sum on the right still has x terms. Thus 2x=x and hence 2=1.

With regard to epsilon-delta arguments remember that lines of a proof do not need justification. They just needs to be true. If your proof starts with the statement 0<1 there is no need to agonise over why you started there..........that is your choice.

Consider the following Theorem: -2=2. Proof: -2=2 implies that |-2|=|2| and hence 2=2 which is true. Therefore -2=2. Logic, like water, usually only flows in one direction. Even in the old days, HSC markers were brutal on responses which ran in the wrong direction, particularly inequality...

Yes What I meant was that there was only one point on the graph where a tangent exists...original corrected

Is it possible for a function to be defined over the entire real line and to have only one point on the graph where a tangent exists (corrected)