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 from Einstein. "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.

Harder Mathcounts State/AMC Questions

2012 Mathcounts State Sprint #30: In rectangle ABCD, shown here, point M is the midpoint of side BC, and point N lies on CD such that DN:NC = 1:4. Segment BN intersects AM and AC at points R and S, respectively. If NS:SR:RB = x:y:z, where x, y and z are positive integers, what is the minimum possible value of x + y + z? 

Solution I :

\(\overline {AB}:\overline {NC}=5:4\) [given]

Triangle ASB is similar to triangle CSN (AAA)

\(\overline {NS}:\overline {SB}= 4 : 5\)

Let \(\overline {NS}= 4a,  \overline {SB}= 5a.\)

Draw a parallel line to \(\overline {NC}\) from M and mark the interception to \(\overline {BN}\)as T.

 \(\overline {MT}: \overline {NC}\) = 1 to 2. [\(\Delta BMT\) and \(\Delta BCN\) are similar triangles ]

\(\overline {NT} = \overline {TB}= \dfrac {4a+5a} {2}=4.5a\)

\(\overline {ST} = 0.5a\)

 \(\overline {MT} :  \overline {AB}\) = 2 to 5
[Previously we know  \(\overline {MT}: \overline {NC}\) = 1 to 2 or 2 to 4 and  \(\overline {NC}:\overline {AB}= 4 : 5\) so the ratio of the two lines  \(\overline {MT} :  \overline {AB}\) is 2 to 5.]


\(\overline {TB} = 4.5 a\)  [from previous conclusion]

Using 5 to 2 line ratio [similar triangles \(\Delta ARB\) and \(\Delta MRT\) , you get \(\overline {BR} =\dfrac {5} {7}\times 4.5a =\dfrac {22.5a} {7}\) and \(\overline {RT} =\dfrac {2} {7}\times 4.5a =\dfrac {9a} {7}\)

Thus, x : y : z = 4a : \( \dfrac {1} {2}a + \dfrac {9a} {7}\) : \(\dfrac {22.5a} {7}\) = 56 : 25 : 45

x + y + z = 126

Solution II : 
From Mathcounts Mini: Similar Triangles and Proportional Reasoning

Solution III: 
Using similar triangles ARB and CRN , you have \(\dfrac {x} {y+z}=\dfrac {5} {9}\).
9x = 5y + 5z ---- equation I

Using similar triangles ASB and CSN and you have \( \dfrac {x+y} {z}=\dfrac {5} {4}\).
4x + 4y = 5z  ---- equation II

Plug in (4x + 4y) for 5z on equation I and you have 9x = 5y + (4x + 4y) ; 5x = 9y ; x = \(\dfrac {9} {5}y\)
Plug in x = \(\dfrac {9} {5}y\) to equation II and you have z  =  \( \dfrac {56} {25}y\)

x : y : z = \(\dfrac {9} {5}y\)  : y  :  \( \dfrac {56} {25}y\) =  45 y :  25y  :  56y

45 + 25 + 56 = 126

Solution IV : Yes, there is another way that I've found even faster, saved for my private students. :D 

Solution V : from Abhinav, one of my students solving another similar question : 

Two other similar questions from 2016 AMC A, B tests : 

2016 AMC 10 A, #19 : Solution from Abhinav 

2016 AMC 10 B #19 : Solution from Abhinav