Law of cosines, without the angle I recently twice on the AMCs came across this problem solving technique for finding side lengths in a triangle with a cevian.  (A cevian is a segment from one vertex of a triangle to the opposite side.)  It uses the law of cosines, but the good news is it may not require an angle measure because it takes advantage of the fact that $cos(180-\alpha)=-cos(\alpha)$.  Note in the diagram that the cevian creates these 2 angles.  By the law cosines $a^2=c^2+e^2-2abcos(\alpha)$ and $b^2=d^2+e^2-2abcos(180-\alpha) = d^2+e^2+2abcos(\alpha)$.  Both equations have the length of the cevian, $e$, and the angle $\alpha$, so you can eliminate a variable and solve.

Bug on a surface

I recently reviewed a problem involving a bug on a cylinder.  We need to find the shortest distance from the bug’s original position, A, to B which is on the opposite side of the cylinder.  The bug can only crawl on the surface, and we’re given the height (8) and circumference (12).  One student wanted the bug to crawl along the lower circumference and then crawl straight up the side.

You can more easily visualize the shortest path by using your imagination to “cut” the cylinder along one side starting at point A, then “flatten” the cylinder into a 12×8 rectangle.  Now the shortest path is easily seen as the hypotenuse of the triangle with legs of 6 and 8.  (A 3-4-5 triangle hiding in plain sight!) Other friendly triangles

Most students are familiar with triangles whose sides are Pythagorean triples like 3-4-5 and 5-12-13.  They may also want to add the 13-14-15 triangle to their list.  It contains both a 5-12-13 triangle and a 3-4-5 triangle (disguised as a 9-12-15, and sharing the altitude of length 12).  This triangle appears often on the AMCs and you’ll save yourself some time if you can recognize it early. Angles on a clockface I worked this problem with some 5th graders last year.  Since I was unsure about their exposure to angles, I began by asking if they knew the degree measure of a right angle, from 12 to 3 on the clock.  From there, I told them I would divide up that 90 degrees into 3 sections, drawing in the 1 and 2 on the clock.  How many degrees between the 12 and 1?

Then I followed the solution provided, and continued by asking them for the degree measure of the angle between 12 and 8, which is the degrees between the 2 hands at 8:00.  Then we stepped through how many degrees we lose by moving the minute hand from the 12 to between the 5 and 6 after 24 minutes.  Then we need to add the few degrees the hour hand moved in 24 minutes.

This approach reminded me that there are two approaches to teaching math.  One way is to repeat the same problem over and over, which little change until the algorithm in memorized.  Another more fun way is to begin with an elementary problem, and do only a small number of those before immediately continuing to progressively more difficult problems.

After I asked them how many degrees were in a right angle, the students then unprompted told me the degree measure from 12 to 6 and then 12 to 9, pretty much by rote.  But I didn’t want to hear their answers to elementary problems.  I wanted to quickly propel them to more difficult problems: the degree measure from 12 to 8 and then at 8:24.  I think they had fun solving this problem and internalized the concept better by solving 1 more difficult problem than a page full of identical easier problems.

(Source: Math Olympiad for Elementary and Middle School)

Use abstraction to sort through the riff-raff

If/then statements can be replaced with $p\rightarrow q$ (read: “p implies q”).  Such a statement is equivalent to its contrapositive ( $\sim q\rightarrow \sim p$, read: “not q implies not p”).  Abstracting the statements by replacing them with variables makes it easier to quickly identify the contrapositive among the statements. Just do something

When you don’t know how to approach a problem, you won’t get anywhere unless you try something, anything.  In this case, find the digit sum of the product when k=1, k=2, k=3 and see if a pattern becomes evident.

(Source:  AMC) Multiples of 11

Every once in a while you’ll need to test a number for divisibility by 11.  Multiples of 11 can be quickly identified if the alternating sum of their digits is a multiple of 11.  This works because the consecutive powers of 10 are either one more or one less than a multiple of 11.  For example, 10 is one less than a multiple of 11, and 100 is one more than a multiple of 11, and 1000 is one less than a multiple of 11.

If you pull out these multiples of 11 from a number, then you are left with an alternating sum.  Since the sum of 2 multiples of 11 is a multiple of 11, then the alternating sum is also a multiple of 11.

This is clearer in my notes, where a, b, c, and d are the digits of a 3 and 4 digit number: 