Friday, December 2, 2016

Grab bag of problems

Sadly, I  have to miss Math club on Tue. given a family emergency. In the meantime, these are some of the problems I've enjoyed looking at recently that I might use sometime soon:

I've reached that point when I see something like this that I think: huh if I form a  right triangle using B to H as the hypotenuse, the remainder of one side is 3/5 r and the hypotenuse is r to the other side must be 4/5 r so its a 3-4-5 triangle.  There are a 1:3 and 1:2 triangle embedded on the edges of the circles as well.  To actually prove this is true just takes the Pythagorean theorem in my version but I suspect there are several other means to the same end. I'd probably scaffold this a bit.

A lovely factoring and deduction exercise. Not too hard beyond the large numbers.  Actually what might be fun is to adapt this and do it interactively in club.

You only need to find the ratio in one of the eight wedges (I drew in ABU) to find the total answer.
Also you could draw a different wedge by extending the inner triangle IJU but I find the first form a bit easier to reason about. Note: the 45-45-90 triangles and the other similar ones. The answer can be found through a combination of similar triangles and area addition starting from the 45-45-90.

The kids actually did this one. It was my second favorite to discuss. My first solution was a rather tedious exercise in the Pythagorean Theorem. There is a similar triangle version that is much simpler.

Some lovely divisibility rule practice embedded in this puzzle.

Find the smallest possible value of the expression:

$$\sqrt{a^2 + 9} + \sqrt{(b-a)^2 + 4} + \sqrt{(8 - b)^2 + 16}$$

This looks complex but ultimately boils down to the shortest distance between n points is a line! Unfortunately it needs a tiny bit of algebra to recognize the geometric representation and find the line so I'll have to save it for some time in the future.

No comments:

Post a Comment