A good formula to always know is that the area of the largest rectangle able to be inscribed within a triangle is always half the area of the triangle.
Excelent idea, thank you ! I think it would have helped if you had assigned letters A,B,C to the vertices of the triangle; D, E, F, G to the vertices of the inscribed rectangle; I, J to the other two points on the base resulted from flipping at 08:40; H to the intersection of those flipped sides
Area of triangle using heron's formula = sqrt [ s ( s - a ) ( s - b ) ( s - c ) ] , where s = ( a + b + c ) / 2 so s = 24 and Area = sqrt [ 24*14*7*3 ] = 84 84 = ( 1 / 2 ) * 21 * h (84 / 21) * 2 = h h = 8
By algebra we can get the same answer. Say the triangle has length a,b and c with one side of the rectangle with length x on c, and the other side of the rectangle equals d, the height of the triangle on side c equals h, x=c(h-d)/h, S=c(h-d)/h*d/2 =c/h((h-d)h/2). c/h is constant, so S is the biggest when (h-d)h is the biggest. (h-d)h=-(d-h/2)^2+1/4*h^2, so the rectangle is the biggest when d=h/2
...A splendid demo of a concept of calculus...the thinner and thiner rectanlges, then adjusting their sizes to show why a particular shape can have maximum area. Are MathCounts {tm} students axquainted with Heron's Formula?
I found the other guy finding the area of a rectangle with sides 6.9 and 10 with calculus. He used coordinate geometry and got a limit as n→infinity of something and got 69 the long way.
Assuming a = the number of horizontal lines and b = the number of vertical lines, then the formula is aC2 * bC2 (C means combination). In a n x m grid, that would have n+1 vertical lines, and m+1 horizontal lines, so the number of rectangles would be ((n+1)C2) * ((m+1)C2). Basically, you are picking two horizontal lines for sides of the rectangles and two vertical lines to be the other two sides of the rectangle. Note: This formula works best for grids, but if dots/points are given instead, in addition to this formula, you have to count all the diagonal rectangles.
A good formula to always know is that the area of the largest rectangle able to be inscribed within a triangle is always half the area of the triangle.
Congratualations you understood the purpose of the video
Excelent idea, thank you ! I think it would have helped if you had assigned letters A,B,C to the vertices of the triangle; D, E, F, G to the vertices of the inscribed rectangle; I, J to the other two points on the base resulted from flipping at 08:40; H to the intersection of those flipped sides
Area of triangle using heron's formula = sqrt [ s ( s - a ) ( s - b ) ( s - c ) ] , where s = ( a + b + c ) / 2
so s = 24 and Area = sqrt [ 24*14*7*3 ] = 84
84 = ( 1 / 2 ) * 21 * h
(84 / 21) * 2 = h
h = 8
That was a really clever way to show that the rectangle was the largest possible
By algebra we can get the same answer. Say the triangle has length a,b and c with one side of the rectangle with length x on c, and the other side of the rectangle equals d, the height of the triangle on side c equals h, x=c(h-d)/h, S=c(h-d)/h*d/2 =c/h((h-d)h/2). c/h is constant, so S is the biggest when (h-d)h is the biggest.
(h-d)h=-(d-h/2)^2+1/4*h^2, so the rectangle is the biggest when d=h/2
Yep
How can we use Heron’s formula in this? I see a lot of comments about this.....
Really elegant solution, I love it. :)
How do you know that if the side is too short that the top piece will go outside of the triangle?
...A splendid demo of a concept of calculus...the thinner and thiner rectanlges, then adjusting their sizes to show why a particular shape can have maximum area. Are MathCounts {tm} students axquainted with Heron's Formula?
jwm239 yep if they are decent
You can use Heron's Formula.
i thought you said Heroine's Formula I was like WOAH
How do you know if one side is smaller that the area will be smaller? Couldn't the other side affect the area as well?
The other side isn’t increasing as fast as the first is decreasing
where is Harvey?????
I used Heron's formula but I accidentally used the full perimeter instead of the semiperimeter in the problem. Whoops! :)
I found the other guy finding the area of a rectangle with sides 6.9 and 10 with calculus.
He used coordinate geometry and got a limit as n→infinity of something and got 69 the long way.
How would you do this all(go through the though process) in the sprint time period?
They probably wouldn't put it in SPRINT.
That is HARD.
@@Chris-wz9jg Not REALLY that hard
@@Chris-wz9jg They would.I've seen Much harder things than those problems on sprint.
@@Chris-wz9jg
The proof part should be automatic. The solving part should take like a minute or two at the max.
I used an alterative method consisting of a derivative to find maximum.
Me be like 5:12
can some plz tell me how to derive the formula for Counting Rectangles in an n X m figure.
Assuming a = the number of horizontal lines and b = the number of vertical lines, then the formula is aC2 * bC2 (C means combination). In a n x m grid, that would have n+1 vertical lines, and m+1 horizontal lines, so the number of rectangles would be ((n+1)C2) * ((m+1)C2). Basically, you are picking two horizontal lines for sides of the rectangles and two vertical lines to be the other two sides of the rectangle. Note: This formula works best for grids, but if dots/points are given instead, in addition to this formula, you have to count all the diagonal rectangles.
What happened in the end? The last sentence after the solution.
Richard Xu inside joke about richard's imaginary friend named harvey who "does his geometry for him".
Dude...i loved it
use heron's formula
If the triangle is not obtuse, the three cases will result in the same answer, area of the triangle/2.
there's a much faster method to find area of triangle, a rectangle minus the 6-8-10 and 5-8-17 triangles.
Or you could've just used Heron's Formula. :P A whole lot easier IMO.
yea, would be alot easier lol
USAMTS shirt! :)
heron
42,the answer's always 42
I see, you are a man of culture as well.
Captain America