Show Your Work, Or, How My Math Abilities Started to Decline

Show your work, or, how my math abilities started to decline

I think it's problematic the way schools teach Algebra. Those meaningless show-your-work approaches, without knowing what Algebra is truly about. The overuse of calculators and the piecemeal way of teaching without the unification of the math concepts are detrimental to our children's ability to think critically and logically.

Of course eventually, it would be beneficial to students if they show their work with the much more challenging word problems (harder Mathcounts state team round, counting and probability questions, etc...), but it's totally different from what some schools ask of our capable students.

How do you improve problem solving skills with tons of worksheets by going through 50 to 100 problems all look very much the same? It's called busy work. 

Quote:  "Insanity: doing the same thing over and over again and expecting different results."

Quotes from Richard Feynman, the famous late Nobel-laureate physicist. Feynman relates his cousin's unhappy experience with algebra:

My cousin at that time—who was three years older—was in high school and was having considerable difficulty with his algebra. I was allowed to sit in the corner while the tutor tried to teach my cousin algebra. I said to my cousin then, "What are you trying to do?" I hear him talking about x, you know."Well, you know, 2x + 7 is equal to 15," he said, "and I'm trying to figure out what x is," and I say, "You mean 4." He says, "Yeah, but you did it by arithmetic. You have to do it by algebra."And that's why my cousin was never able to do algebra, because he didn't understand how he was supposed to do it. I learned algebra, fortunately, by—not going to school—by knowing the whole idea was to find out what x was and it didn't make any difference how you did it. There's no such a thing as, you know, do it by arithmetic, or you do it by algebra. It was a false thing that they had invented in school, so that the children who have to study algebra can all pass it. They had invented a set of rules, which if you followed them without thinking, could produce the answer. Subtract 7 from both sides. If you have a multiplier, divide both sides by the multiplier. And so on. A series of steps by which you could get the answer if you didn't understand what you were trying to do.
So I was lucky. I always learnt things by myself.

The Grid Technique in Solving Harder Mathcounts Counting Problems : from Vinjai

The following notes are from Vinjai, a student I met online. He graciously shares and offers the tips here on how to tackle those harder Mathcounts counting problems. 

The point of the grid is to create a bijection in a problem that makes it easier to solve. Since the grid just represents a combination, it can be adapted to work with any problem whose answer is a combination.

For example, take an instance of the classic 'stars and bars' problem (also known as 'balls and urns', 'sticks and stones', etc.):

Q: How many ways are there to pick an ordered triple (a, b, c) of nonnegative integers such that a+b+c = 8? (The answer is 10C2 or 45 ways.)

Solution I: 

This problem is traditionally solved by thinking of ordering 8 stars and 2 bars. An example is:

* * * |    | * * * * *

  ^       ^       ^

  a       b       c

This corresponds to a = 3, b = 0, c = 5.

Solution II: 

But this can also be done using the grid technique:

The red path corresponds to the same arrangement: a = 3, b = 0, c = 5. The increase corresponds to the value: a goes from 0 to 3 (that is an increase of 3), b goes from 3 to 3 (that is an increase of 0), and c goes from 3 to 8 (that is an increase of 5). So a = 3, b = 0, c = 5.

Likewise, using a clever 1-1 correspondence, you can map practically any problem with an answer of nCk to fit the grid method. The major advantage of this is that it is an easier way to think about the problem (just like the example I gave may be easier to follow than the original stars and bars approach, and the example I gave in class with the dice can also be thought of in a more numerical sense).

Similar Triangles: Team question : Beginning level

9. In the figure below, quadrilateral CDEG is a square with CD = 3, and quadrilateral BEFH is a rectangle. If EB = 5, how many units is BH? Express your answer as a mixed number

Triangle BED is a 3-4-5 right triangle and is similar to triangle GEF.

BE : ED = GE : EF = 5 : 3 = 3 : FE

EF = 9/5 = BH  The answer

Some articles on problem solving and parenting

Why America's Smartest Students Fail Math 

Are our kids failing in math because they can't read ? 

Kids of Helicopter Parents Are Sputtering Out 

The Juilliard Effect : Ten Years Later 

Ay reflection notes

Ay

 5/7 

1) Reviewed the attached problems and solutions and I understand them well

2) Attempted the 2024 AMC 12A in 60 minutes. I attempted #1 to #19 and got them all correct.

I had a headache after that so I couldn't attempt the rest of the questions. Skipped  -#20 to #25

Not sure how many I could have actually done.