January 31, 2012
Free ebook "All Children Can Be Great Listeners"
January 30, 2012
Triangle problem with equal areas - solution
This is the solution for the triangle problem with equal areas that I posted earlier.
We are given that the areas of the three right triangles are equal, that is the area of the triangle DAE = area of the triangle EBF = area of the triangle FCD.
We will make an equation based on that fact. For that, I like to use x as my variable, so I denote the longer side of the rectangle with a, the other side with b, the distance AE with x, and the distance BF with y. We are asked the ratio AE:EB, which is the same as x : (a − x) using my notation, and the ratio BF:FC, which is the same as y : (b − y) using my notation.
The area of triangle ADE is its base times altitude divided by 2, or bx/2.
The area of triangle EBF is its base times altitude divided by 2, or y(a − x)/2.
The area of triangle CDF is its base times altitude divided by 2, or a(b − y)/2.
And these three are equal. Basically you just make two equations from the above information, and manipulate your equations until you get the ratios expressed without x and y.
Since the area of triangle ADE = area of triangle EBF, we get:
bx = y(a − x)
bx = ay − yx
Since the area of triangle EBF = area of triangle CDF, we get:
y(a − x) = a(b − y)
ay − yx = ab − ay
Now I have two equations with two unknowns (x and y) to solve, or a system of equations:
bx = ay − yx
ay − yx = ab − ay
We'll solve them, and then write the ratios x : (a − x) and y : (b − y).
========================================================
Well, to save time... I did solve these on paper the other day but lost it, and writing it in html is not my favorite hobby... so I just put these equations into Wolframalpha, separated by a semicolon, and got
x = (3a - √5a)/2, y = (√5 − 1)b / 2
OR
x = (3a + √5a)/2, y = (√5 − 1)b / 2
Then the two ratios. We need to substitute the value of x from above into the ratio x : (a − x). I'll use the first solution first.
x : (a − x) = x / (a − x)
= ((3a - √5 a)/2 ) / a − (3a - √5 a)/2)
Hey, maybe Wolframalpha will simplify this expression for me, too. And yes, it did. The simplified form is 1/2 (√5 - 1).
What happens if I use the second solution for x?
In the ratio x / (a − x), we get
(3a + √5a)/2 /(a − ((3a + √5a)/2))
The a's will cancel out from this expression, so we get
(3a + √5a)/2 /(a − ((3a + √5a)/2))
Simplifying with Wolframalpha again, I get (−1 − √5)/2 which is a negative number. So I discard that solution.
Then on to the second ratio, y : (b − y), using the (only) solution y = (√5 − 1)b / 2 first. I substitute that and get ((√5 − 1)b / 2) / (b − ((√5 − 1)b / 2)).
Again, b's cancel out so we have ((√5 − 1)/2) / (1 − ((√5 − 1)/ 2))
Simplifying that, we get (1 + √5)/2, which is the GOLDEN RATIO (or Phi).
So, the two ratios are (√5 + 1)/2 and (√5 − 1)/2, or the golden ratio and its conjugate. Notice these don't depend on a or b (the sides of the rectangle) — meaning that these ratios are the same in any rectangle.
We are given that the areas of the three right triangles are equal, that is the area of the triangle DAE = area of the triangle EBF = area of the triangle FCD.
We will make an equation based on that fact. For that, I like to use x as my variable, so I denote the longer side of the rectangle with a, the other side with b, the distance AE with x, and the distance BF with y. We are asked the ratio AE:EB, which is the same as x : (a − x) using my notation, and the ratio BF:FC, which is the same as y : (b − y) using my notation.
The area of triangle ADE is its base times altitude divided by 2, or bx/2.
The area of triangle EBF is its base times altitude divided by 2, or y(a − x)/2.
The area of triangle CDF is its base times altitude divided by 2, or a(b − y)/2.
And these three are equal. Basically you just make two equations from the above information, and manipulate your equations until you get the ratios expressed without x and y.
Since the area of triangle ADE = area of triangle EBF, we get:
| bx/2 | = | y(a − x)/2 |
bx = y(a − x)
bx = ay − yx
Since the area of triangle EBF = area of triangle CDF, we get:
| y(a − x)/2 | = | a(b − y)/2 |
y(a − x) = a(b − y)
ay − yx = ab − ay
Now I have two equations with two unknowns (x and y) to solve, or a system of equations:
bx = ay − yx
ay − yx = ab − ay
We'll solve them, and then write the ratios x : (a − x) and y : (b − y).
========================================================
Well, to save time... I did solve these on paper the other day but lost it, and writing it in html is not my favorite hobby... so I just put these equations into Wolframalpha, separated by a semicolon, and got
x = (3a - √5a)/2, y = (√5 − 1)b / 2
OR
x = (3a + √5a)/2, y = (√5 − 1)b / 2
Then the two ratios. We need to substitute the value of x from above into the ratio x : (a − x). I'll use the first solution first.
x : (a − x) = x / (a − x)
= ((3a - √5 a)/2 ) / a − (3a - √5 a)/2)
Hey, maybe Wolframalpha will simplify this expression for me, too. And yes, it did. The simplified form is 1/2 (√5 - 1).
What happens if I use the second solution for x?
In the ratio x / (a − x), we get
(3a + √5a)/2 /(a − ((3a + √5a)/2))
The a's will cancel out from this expression, so we get
(3a + √5a)/2 /(a − ((3a + √5a)/2))
Simplifying with Wolframalpha again, I get (−1 − √5)/2 which is a negative number. So I discard that solution.
Then on to the second ratio, y : (b − y), using the (only) solution y = (√5 − 1)b / 2 first. I substitute that and get ((√5 − 1)b / 2) / (b − ((√5 − 1)b / 2)).
Again, b's cancel out so we have ((√5 − 1)/2) / (1 − ((√5 − 1)/ 2))
Simplifying that, we get (1 + √5)/2, which is the GOLDEN RATIO (or Phi).
So, the two ratios are (√5 + 1)/2 and (√5 − 1)/2, or the golden ratio and its conjugate. Notice these don't depend on a or b (the sides of the rectangle) — meaning that these ratios are the same in any rectangle.
January 28, 2012
Bon from Math is not a four-letter word made this little counting song, sung to the tune of "Twinkle Twinkle Little Star". Hope your little ones enjoy it!
January 24, 2012
Three-day sale for Math Mammoth
SORRY I forgot to post it here (I just sent this to my email list).
For January 23-25, get 23% off of all Math Mammoth downloads & CDs
at Kagi store.
Use the coupon code THREEDAYS.
Go to http://www.mathmammoth.com first, then find the links to
Kagi's order pages. OR, use these direct links:
~ Light Blue series (complete curriculum)
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=Math_Mammoth_LightBlue_Series
~ Blue series
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE
~ Golden and Green Series
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=MathMammoth_Workbooks
~ Make It Real Learning workbooks
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=Make_It_Real_Learning
~ Bundles (CDs or downloads)
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=Math_Mammoth_Packages
THREE DAYS ONLY!
For January 23-25, get 23% off of all Math Mammoth downloads & CDs
at Kagi store.
Use the coupon code THREEDAYS.
Go to http://www.mathmammoth.com first, then find the links to
Kagi's order pages. OR, use these direct links:
~ Light Blue series (complete curriculum)
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=Math_Mammoth_LightBlue_Series
~ Blue series
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE
~ Golden and Green Series
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=MathMammoth_Workbooks
~ Make It Real Learning workbooks
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=Make_It_Real_Learning
~ Bundles (CDs or downloads)
https://store.kagi.com/cgi-bin/store.cgi?storeID=5KN_LIVE&page=Math_Mammoth_Packages
THREE DAYS ONLY!
January 23, 2012
Triangle puzzle - equal areas
I hope Pat doesn't mind that I copied the image from his blog... He posted this triangle puzzle on his blog and I thought you might enjoy it, too!
Basically, we have a triangle DFE inside a rectangle, dividing the rectangle into various triangles. And, the three areas of fainter color are equal. That is, the area of the triangle DAE = area of the triangle EBF = area of the triangle FCD. (Notice the image is not drawn to scale at all.)
And, we're asked to solve the RATIOS AE : EB and BF : FC.
I'll post a solution later.
And, we're asked to solve the RATIOS AE : EB and BF : FC.
I'll post a solution later.
Versa Ruler
Well, this IS something different! A physical ruler that can draw shapes with any angle measure you want.
Unfortunately it's not yet in production. In fact, the project is needing funding... but very interesting! Please read more at
Rule Like Never Before! NEW Shape-making Versa Ruler
This ruler give you accurate measurements for every side and angle. You can connect sides to form angles, which scale and skew smoothly, and you can lock angles and sides. Cool!
Unfortunately it's not yet in production. In fact, the project is needing funding... but very interesting! Please read more at
Rule Like Never Before! NEW Shape-making Versa Ruler
This ruler give you accurate measurements for every side and angle. You can connect sides to form angles, which scale and skew smoothly, and you can lock angles and sides. Cool!
January 21, 2012
Vi Hart and mathematical doodling
I just learned about Vi Hart's "doodling in math class" videos (hat tip goes to Fawn Nguyen).
Vi calls herself mathemusician - and definitely, she's a talented and smart gal! I'm sure you'll enjoy her videos (as long as you can follow her super fast speeaking). Here are some that I enjoyed:
Doodling in Math: Spirals, Fibonacci, and Being a Plant [1 of 3]
Here's a link to learn more about Fibonacci numbers in nature, so you can read about it at a slow pace : )
Binary Hand Dance was pretty cool too!
Her series of "mathematical doodling" videos have become somewhat of a viral success. Here's one more:
Doodling in Math Class: Infinity Elephants
Vi calls herself mathemusician - and definitely, she's a talented and smart gal! I'm sure you'll enjoy her videos (as long as you can follow her super fast speeaking). Here are some that I enjoyed:
Doodling in Math: Spirals, Fibonacci, and Being a Plant [1 of 3]
Here's a link to learn more about Fibonacci numbers in nature, so you can read about it at a slow pace : )
Binary Hand Dance was pretty cool too!
Her series of "mathematical doodling" videos have become somewhat of a viral success. Here's one more:
Doodling in Math Class: Infinity Elephants
January 20, 2012
Math teachers are at play again
Denise's done a beautiful job with the current Math Teachers at Play carnival number 46, lots to read and explore and see... head on over!
January 09, 2012
Compound interest
Someone sent me in a question concerning compound interest...
The formula that this person is using is correct... the formula for compound interest is
A = p(1 + r)t
but this formula doesn't give us the amount of interest -- it gives us the amount of money you would withdraw after t years. In the formula, p is the original principal, r is the interest rate, and t is the time in years.
However, we cannot put the interest rate in as he did. If r = 10%, then r = 0.1 must be used in this formula. In other words, FIRST convert your percentage into a decimal.
For example, if the principal is $5000 and r = 10% = 0.1, then we get
A = $5000 × 1.1t
The number 1 in the formula p(1 + r)t doesn't stand for anything by itself. It comes from using the distributive property in simplifying p + pr into p(1 + r).
Which, p + pr is the principal + interest earned after one year; that is what you could withdraw after one year.
BUT if you don't withdraw it, but leave it all to earn more interest, then that becomes your "new" principal, and the year after that you will have that times 1 + r.
Basically, this is how compound interest works:
After year 1: You have p + pr which is p(1 + r). Notice that your original principal got multiplied by (1 + r). If you leave this on the account to earn more interest, then next year you have that amount times (1 + r).
After year 2: you have p(1 + r)(1 + r). If you leave this on the account to earn more interest, then next year you have that amount times (1 + r).
After year 3: you have p(1 + r)(1 + r)(1 + r). Let's simplify this using an exponent: You have p(1 + r)3
After year 4: you have p(1 + r)4
And so on. After year t, you have p(1 + r)t.
Hope this helps some!
Please send me the formula for compound interest and explain line by line.
p(1 + rate)3
What does the 1 stand for and must you add it to the rate of say 10%?
100(1+10)3 years = ???
Obviously I am looking for a basic course?
The formula that this person is using is correct... the formula for compound interest is
but this formula doesn't give us the amount of interest -- it gives us the amount of money you would withdraw after t years. In the formula, p is the original principal, r is the interest rate, and t is the time in years.
However, we cannot put the interest rate in as he did. If r = 10%, then r = 0.1 must be used in this formula. In other words, FIRST convert your percentage into a decimal.
For example, if the principal is $5000 and r = 10% = 0.1, then we get
A = $5000 × 1.1t
The number 1 in the formula p(1 + r)t doesn't stand for anything by itself. It comes from using the distributive property in simplifying p + pr into p(1 + r).
Which, p + pr is the principal + interest earned after one year; that is what you could withdraw after one year.
BUT if you don't withdraw it, but leave it all to earn more interest, then that becomes your "new" principal, and the year after that you will have that times 1 + r.
Basically, this is how compound interest works:
After year 1: You have p + pr which is p(1 + r). Notice that your original principal got multiplied by (1 + r). If you leave this on the account to earn more interest, then next year you have that amount times (1 + r).
After year 2: you have p(1 + r)(1 + r). If you leave this on the account to earn more interest, then next year you have that amount times (1 + r).
After year 3: you have p(1 + r)(1 + r)(1 + r). Let's simplify this using an exponent: You have p(1 + r)3
After year 4: you have p(1 + r)4
And so on. After year t, you have p(1 + r)t.
Hope this helps some!
January 02, 2012
Math Mammoth & Common Core Standards - announcement
I will be aligning Math Mammoth to the Common Core standards. This might take me half a year. Thus far it looks like just minor changes for grades 1-5, and a bit more changes for grade 6.
Nearly all US states (only 5 or 6 haven't) either have adopted or will adopt the Common Core standards, so this means that Math Mammoth will probably be aligned to YOUR state's standards as well!
Actually, Math Mammoth is even now fairly well aligned to the Common Core standards. Common Core standards are based on having a few "focus" areas for each grade level (instead of the "inch deep and mile wide" curriculum so prevalent in the past), and Math Mammoth being a mastery-based program has always had similar basic focus areas for each grade.
Nearly all US states (only 5 or 6 haven't) either have adopted or will adopt the Common Core standards, so this means that Math Mammoth will probably be aligned to YOUR state's standards as well!
Actually, Math Mammoth is even now fairly well aligned to the Common Core standards. Common Core standards are based on having a few "focus" areas for each grade level (instead of the "inch deep and mile wide" curriculum so prevalent in the past), and Math Mammoth being a mastery-based program has always had similar basic focus areas for each grade.
December 27, 2011
Math Mammoth books and their origin - video
Here's a little 2-minute video of me talking about Math Mammoth books, their origin, and also featuring "Mathy" the mammoth. : )
Subscribe to:
Posts (Atom)