I overcomplicated this and was looking at solving the side of the hexagon given the area and looking into heights and widths of the hexagons to find the x & y of the rectangle and subtract 12... This is a much more elegant solution!
I initially overcomplicated it too on my first attempt. Then I took a step back and realized that I don't have to worry about the half-triangles on each side. since there are two half-triangles on each side, they could be counted as 1 equilateral triangle. I also realized the rectangle can be split into 3 rows. Each row has a area of 7, and the total area of the rectangle is 21. When I subtract 12 from 21, you get 9. The formula would be: (7*3)-(6*2)=9
You've really been cranking these out, and I've been following along the entire month. I love the videos, and I appreciate the effort you've been putting into them, thanks for making these!
@@jeffdlong Geometry is a part of mathematics. Since it is a regular hexagon meaning all sides and angles are the same, using the formula for sum of angles in a convex polygon (n-2)*180° based on number, n, of sides, each angle will be (6-2)*180°/6=120°. Any line segments constructed within a regular polygon from one of its vertices that creates two halves, i.e. equal area polygons, must be angle bisectors at the vertex, and the midpoint of this line segment is the center of the regular polygon, so drawing three such line segments within the regular hexagon will create six triangles with two 60° angles, therefore the third angle must also be 60°, since the sum of angles in a triangle is (3-2)*180°=180°. The "tiling" of the rest of the rectangle, follows from using equilateral triangles with area of 1. All the triangles with area 1/2 occur because the rectangle boundaries cut the equilateral triangle in half, which, since an equilateral triangle is a regular polygon, then as explained previously means by definition of an angle bisector or (1/2)*60°=30° at the vertex and one 60° angle still exists, so the other must be 90° (which can easily be confirmed at two of the corners of the rectangle).
@@jeffdlong Geometry is a part of mathematics. Since it is a regular hexagon meaning all sides and angles are the same, using the formula for sum of angles in a convex polygon (n-2)*180° based on number, n, of sides, each angle will be (6-2)*180°/6=120°. Any line segments constructed within a regular polygon from one of its vertices that creates two halves, i.e. equal area polygons, must be angle bisectors at the vertex, and the midpoint of this line segment is the center of the regular polygon, so drawing three such line segments within the regular hexagon will create six triangles with two 60° angles, therefore the third angle must also be 60°, since the sum of angles in a triangle is (3-2)*180°=180°. The "tiling" of the rest of the rectangle, follows from using equilateral triangles with area of 1. All the triangles with area 1/2 occur because the rectangle boundaries cut the equilateral triangle in half, which, since an equilateral triangle is a regular polygon, then as explained previously means by definition of an angle bisector or (1/2)*60°=30° at the vertex and one 60° angle still exists, so the other must be 90° (which can easily be verified at two of the corners of the rectangle).
Geometry is a part of mathematics. Since it is a regular hexagon, by definition, all sides are of equal length and all angles are equal in measure. Using the formula for sum of angles in a convex polygon (n-2)*180° based on the number, n, of sides, each angle will be (6-2)*180°/6=120°. Any line segments constructed within a regular polygon from one of its vertices that creates congruent halves of the polygon must be angle bisectors at the vertex, and the midpoints of these line segments coincide since they would be diameters of a circumscribed circle, i.e. a circle upon which all vertices of the polygon lie, so constructing three such line segments within the regular hexagon will create six triangles with two 60° angles, therefore the third angle must also be 60°, since the sum of angles in a triangle is (3-2)*180°=180°. The "tiling" of the rest of the rectangle follows placing equilateral triangles with area of 1 side by side, just as they appear in the hexagon. Since an equilateral triangle is a regular polygon, then the triangles with area 1/2 occur because the rectangle boundaries cut any overlapping equilateral triangles with area 1 in half.
75, I think a triangle with two legs ending at a tangent on a circle is an isosceles triangle, which means both legs are equal, so the hypotenuse(c) is equal to the left side of the rectangle (a) plus the bottom side of the rectangle (b) minus the sum of the legs of the right angle isosceles triangle if you add in lines to the centre of the circle you will see it's a square with the sides of the inscribed angle so a+b-5=c now if you look at the angles and each corner adding up to 90 degrees you can see the three triangles are similar, and from the radii you can see how much different, so you can get the equations so you can see that a=(3c)/5 and b=(4c)/5 so plugging them into the first equation ((3c)/5)+((4c)/5)-5=c which boils down to 12.5 plug this into the second and third equations and you get a=10 and b=7.5 the area of the rectangle=ab=75
Let side of the hexagon has length of a, let height of each triangle making hexagon has length of h. Then a*h/2 == 1, or a*h == 2 for simplification. Let rectangle has height of H and width of L, so the area will be H * L. H is equal to 3 h, L is equal to 3.5 a, because hexagons align that way, which makes the area 3 h * 3.5 * => 10.5 * a * h. Knowing a*h == 2 it makes whole rectangle 21, minus 12 for hexagons it gives 9. There is other way, computing a and h knowing that h is a*sqrt(3) (or going through Pythagorean theorem to get that), and then do computations using units, but it seemed to complicated to do it in 2 minutes.
R1=3/2 R2=4/2 R3=5/2 Okay, since the triangles are similar by inspection of the angles, the inradii must be similar by same ratio (abc~bde~cef, where f=a+d), so a=3/4*b b=3/4*d a=3/5*c b=3/5*e b=4/5*c d=4/5*e Then a=3/4*b=3/5*c=9/16*d=9/20*e By inradius definition, r1=TA1/s1=(ab/2)/((a+b+c)/2) 3/2=(a*4/3*a/2)/((a+4/3*a+5/3*a)/2)=a/3 So a=9/2 b=6 c=15/2 d=8 e=10 And Area=ce=75
The three triangles are similar. Their linear ratios are 3 : 4 : 5 Their hypotenuses correspond to the diagonal, the length and the width of the rectangle. So all three triangles are 3 - 4 - 5 triangles. We can write the length and width and the diagonal of the rectangle as 4x & 3x & 5x, which happen to be the three sides of the large triangle (the one with the red circle inside), we can write 4x - 5/2 + 3x - 5/2 = 5x 2x = 5, x = 5/2 So the rectangle area = (3x)(4x) = 12x^2 = 12(5/2)^2 = 75
Everything is super clear what you said, except the first equation. Where's that come from? Ah, okay! I can see it know. It definitely needs way more explanation though.
@ If you focus on the large triangle, its three sides are 3x (rectangle width), 4x (rectangle length) and 5x (rectangle diagonal), the three sides are all tangent lines of the red circle. The 4x is divided into 2.5 and 4x - 2.5 at the point of tangency (the point of contact between the tangent and the circle). The 3x is divided into 2.5 and 3x - 2.5, the 5x is divided into 4x - 2.5 and 3x - 2.5. Therefore 5x = (4x - 2.5) + (3x - 2.5)
@@Epyxoid Yes, if two lines that are tangent to a circle intersect at a single point, the point is called the "external point of tangency". The distance between the "external point of tangency" and the point of tangency (where the tangent touches the circle circumference) is the same for the two tangents. It is easy to demonstrate, just connect the circle center to the "external point of tangency“, and to the two points of tangency. you now have two identical right triangles that share the hypotenuse, and they both have the radius as one of their side length.
I solved it similarly but using just the side length S, it'd take a very long, but straightforward, proof to show all that even works, but I think showing that all those equilateral triangles fit in that exact shape would be even longer and more tedious work, then again, the equilateral triangle approach gives much faster end calculations so maybe we end up with the same proof length.
Same here. I didn't see enough "rigor" in assuming that the rest of the rectangle had a bunch of equilatoral triangles - it kind of "jumped" to that - but knew we could get the height and width of the rectangle from the triangles themselves.
@@Insightfill You know regular hexagons can be tiled with no gaps, yes? That being the case, all you have to do is look at where the sides and vertices of the two hexagons touch the rectangle and it becomes obvious.
Ooh the next one's easy if you know that a 3-4-5 triangle has an incircle with radius 1. The three triangles must be scaled 3-4-5 so a 3-4-5 triangle is what will work. So the largest triangle has legs 15 and 20, those are the sides of the rectangle, so the rectangle's area is 300 sq u.
@@leolin3207 Oh crap that's the diameter. Yeah, so since the radius is half of what I thought, the area is a quarter of what I thought since (1/2)^2 = 1/4. 75 sq u, final answer Regis.
@@keith6706 Look at how a leg of one plus a leg of another equals the hypotenuse of the third. Since the inradius (or in-diameter in this case) of two similar triangles scales at the same ratio as the sides, the largest triangle has a hypotenuse that's 5/3 the hypotenuse of the smallest one. The math is easy, but since I've seen this configuration a lot I didn't need to scribble it out. I'm sure I;ve mentioned enough to make that path obvious.
You were clean shaven on day 21 puzzle, had 2 to 3 days growth of beard on day 22 puzzle and are clean shaven on day 23. Let’s use a 30 60 90 triangle to see how often Andy shaves his face. 😂😂
Let's assume that the length of the rectangle is a, its width is b, and its diameter is c. We get the equations a+b-c=5, a²+ab-ac=4c, b²+ab-bc=3c. By solving this set of equations, we find a=10, b=15/2, c=25/2. So the area of the rectangle is 10*15/2=75.
Found the answer for tomorrow's puzzle. I knew it was gonna be a sea of equations and tons of algebra work, just didn't know whether or not I'd be able to get every single length, thought maybe I'd just get a L*W = sth, at the end. Anyways, SPOILER FOR THE SOLUTION. The answer I got for the area of the rectangle was 75u². Ok, in order to get that I needed to create from scratch around 5 main equations for 5 unknowns. Now obviously I had a lot of extra equations just in case (only 5 were needed to get my 5 unknowns, not sure which were unnecessary), and obviously I'm gonna write the actual equations here, I'm only gonna write their type, how I got each, the main idea, and how many I got of each. First off, there's a formula for circles inside right triangles (Incircles) that has the inradius in it (radius of the incircle). R = (a + b - c)/2, where a, b, and c, are the sides and c is the hypotenuse. Made three of those for each incircle. We need 2 more equations. I avoided using the equations with Pythagoras theorem, since they'd get me squares and square roots and make the equations much more complicated, so I used similarities. My variables were: the rectangle length, and width, the side from the upper corner to the diagonal, and the 2 pieces it was separated into (which makes 5 unknowns), named: L, W, b, a, and c. Each of those triangle is similar to the 2 others via AA (NOT CONGRUENT THOUGH). 2 similarity equations are more than enough. Now getting the exact length of b was a piece of cake (the common leg of the smallest triangles), I just used the first 3 equations, addion and substitution with the third, and boom, got b = 6 (just by eye observation). Getting the other 4 was the part that needed a lot of work. Some editing to the 5 equations after getting the b, then some algebra got me that, L = 10, W = 7.5
I noticed that you wrote a lot, brother, the issue is not that difficult, and at first, it appears to you that the equations are complicated, and this is not true. To clarify this, let us assume that the length of the rectangle is a, its width is b, and its diameter is c. In each right triangle, we write the equation that connects the sides of the right triangle with the radius of the circle drawn inside it, so we get the following equations: a+b-c=5, a²+ab-ac=4c, b²+ab-bc=3c. We substitute in the second equation b-c=5-a, and substitute in the third equation a-c=5-b, so we get 5a=4c, 5b=3c. Here we get an equation with one unknown c, which is 4c/5+3c/5-c=5, and from it c=25/2, so a=10 and b=15/2, so the area of the rectangle is equal to 10*15/2=75. You can notice from the beginning that the sides of the right triangle are proportional to the diameters of the circles, i.e. a=4x, b=3x, and c=5x, where there is one unknown we are looking for and we find it x=5/2.
There's no assumption. It's insight acquired by experience. We can see that the two hexagons touch the rectangle with two sides and two corners, which means what he did is completely works.
@@rdgspams5669 Geometry is a part of mathematics. Since it is a regular hexagon meaning all sides and angles are the same, using the formula for sum of angles in a convex polygon (n-2)*180° based on number, n, of sides, each angle will be (6-2)*180°/6=120°. Any line segments constructed within a regular polygon from one of its vertices that creates two halves, i.e. equal area polygons, must be angle bisectors at the vertex, and the midpoint of this line segment is the center of the regular polygon, so drawing three such line segments within the regular hexagon will create six triangles with two 60° angles, therefore the third angle must also be 60°, since the sum of angles in a triangle is (3-2)*180°=180°. The "tiling" of the rest of the rectangle, follows from using equilateral triangles with area of 1. All the triangles with area 1/2 occur because the rectangle boundaries cut the equilateral triangle in half, which, since an equilateral triangle is a regular polygon, then as explained previously means by definition of an angle bisector or (1/2)*60°=30° at the vertex and one 60° angle still exists, so the other must be 90° (which can easily be verified at two of the corners of the rectangle).
Geometry is a part of mathematics. Since it is a regular hexagon, by definition, all sides are of equal length and all angles are equal in measure. Using the formula for sum of angles in a convex polygon (n-2)*180° based on the number, n, of sides, each angle will be (6-2)*180°/6=120°. Any line segments constructed within a regular polygon from one of its vertices that creates congruent halves of the polygon must be angle bisectors at the vertex, and the midpoints of these line segments coincide since they would be diameters of a circumscribed circle, i.e. a circle upon which all vertices of the polygon lie, so constructing three such line segments within the regular hexagon will create six triangles with two 60° angles, therefore the third angle must also be 60°, since the sum of angles in a triangle is (3-2)*180°=180°. The "tiling" of the rest of the rectangle follows placing equilateral triangles with area of 1 side by side, just as they appear in the hexagon. Since an equilateral triangle is a regular polygon, then the triangles with area 1/2 occur because the rectangle boundaries cut any overlapping equilateral triangles with area 1 in half
30-60-90 triangle saves the day once more
i cant escape the 30 60 90 triangle
1:49 I loved how the box ate the 30-60-90 triangle
very creative
1:49 Haha, you had to erase the remaining triangle 😂
I overcomplicated this and was looking at solving the side of the hexagon given the area and looking into heights and widths of the hexagons to find the x & y of the rectangle and subtract 12... This is a much more elegant solution!
I did the same thing. I always pick the hard way.
@@baselinesweb Me too, but I failed. What is the side of such a triangle with area 1?
When in doubt, triangles
@@sirllamaiii9708Literally
I initially overcomplicated it too on my first attempt. Then I took a step back and realized that I don't have to worry about the half-triangles on each side. since there are two half-triangles on each side, they could be counted as 1 equilateral triangle. I also realized the rectangle can be split into 3 rows. Each row has a area of 7, and the total area of the rectangle is 21. When I subtract 12 from 21, you get 9.
The formula would be:
(7*3)-(6*2)=9
surprisingly easy for something that looks hard
@@practicemodebutton7559 what’s the area of your practice button?
You've really been cranking these out, and I've been following along the entire month. I love the videos, and I appreciate the effort you've been putting into them, thanks for making these!
It feels incomplete without proving you can tile the rectangle like that. That's the core part of the reasonning.
Agree, i prefer when he uses math to solve
@@jeffdlong
Geometry is a part of mathematics.
Since it is a regular hexagon meaning all sides and angles are the same, using the formula for sum of angles in a convex polygon (n-2)*180° based on number, n, of sides, each angle will be (6-2)*180°/6=120°. Any line segments constructed within a regular polygon from one of its vertices that creates two halves, i.e. equal area polygons, must be angle bisectors at the vertex, and the midpoint of this line segment is the center of the regular polygon, so drawing three such line segments within the regular hexagon will create six triangles with two 60° angles, therefore the third angle must also be 60°, since the sum of angles in a triangle is (3-2)*180°=180°. The "tiling" of the rest of the rectangle, follows from using equilateral triangles with area of 1. All the triangles with area 1/2 occur because the rectangle boundaries cut the equilateral triangle in half, which, since an equilateral triangle is a regular polygon, then as explained previously means by definition of an angle bisector or (1/2)*60°=30° at the vertex and one 60° angle still exists, so the other must be 90° (which can easily be confirmed at two of the corners of the rectangle).
@@jeffdlong
Geometry is a part of mathematics.
Since it is a regular hexagon meaning all sides and angles are the same, using the formula for sum of angles in a convex polygon (n-2)*180° based on number, n, of sides, each angle will be (6-2)*180°/6=120°. Any line segments constructed within a regular polygon from one of its vertices that creates two halves, i.e. equal area polygons, must be angle bisectors at the vertex, and the midpoint of this line segment is the center of the regular polygon, so drawing three such line segments within the regular hexagon will create six triangles with two 60° angles, therefore the third angle must also be 60°, since the sum of angles in a triangle is (3-2)*180°=180°. The "tiling" of the rest of the rectangle, follows from using equilateral triangles with area of 1. All the triangles with area 1/2 occur because the rectangle boundaries cut the equilateral triangle in half, which, since an equilateral triangle is a regular polygon, then as explained previously means by definition of an angle bisector or (1/2)*60°=30° at the vertex and one 60° angle still exists, so the other must be 90° (which can easily be verified at two of the corners of the rectangle).
Geometry is a part of mathematics.
Since it is a regular hexagon, by definition, all sides are of equal length and all angles are equal in measure. Using the formula for sum of angles in a convex polygon (n-2)*180° based on the number, n, of sides, each angle will be (6-2)*180°/6=120°. Any line segments constructed within a regular polygon from one of its vertices that creates congruent halves of the polygon must be angle bisectors at the vertex, and the midpoints of these line segments coincide since they would be diameters of a circumscribed circle, i.e. a circle upon which all vertices of the polygon lie, so constructing three such line segments within the regular hexagon will create six triangles with two 60° angles, therefore the third angle must also be 60°, since the sum of angles in a triangle is (3-2)*180°=180°. The "tiling" of the rest of the rectangle follows placing equilateral triangles with area of 1 side by side, just as they appear in the hexagon. Since an equilateral triangle is a regular polygon, then the triangles with area 1/2 occur because the rectangle boundaries cut any overlapping equilateral triangles with area 1 in half.
He should've left that tiny red piece all the way through the end. People would go crazy about it.
I always watch you solve math problems as part of my morning routine to set off a positive vibe to my day. How exciting.
Easy puzzle AND Andy uses "comprise" correctly = my day made.
I love these videos and no I'm not watching them for school I really enjoy math and theses videos help me explore my passion
75, I think
a triangle with two legs ending at a tangent on a circle is an isosceles triangle, which means both legs are equal, so the hypotenuse(c) is equal to the left side of the rectangle (a) plus the bottom side of the rectangle (b) minus the sum of the legs of the right angle isosceles triangle if you add in lines to the centre of the circle you will see it's a square with the sides of the inscribed angle
so a+b-5=c
now if you look at the angles and each corner adding up to 90 degrees you can see the three triangles are similar, and from the radii you can see how much different, so you can get the equations
so you can see that a=(3c)/5 and b=(4c)/5
so plugging them into the first equation
((3c)/5)+((4c)/5)-5=c
which boils down to 12.5
plug this into the second and third equations and you get
a=10 and b=7.5
the area of the rectangle=ab=75
I think Pi will show up in the next one. Go Andy! Three more days. This has been fun.
Let side of the hexagon has length of a, let height of each triangle making hexagon has length of h. Then a*h/2 == 1, or a*h == 2 for simplification.
Let rectangle has height of H and width of L, so the area will be H * L.
H is equal to 3 h, L is equal to 3.5 a, because hexagons align that way, which makes the area 3 h * 3.5 * => 10.5 * a * h. Knowing a*h == 2 it makes whole rectangle 21, minus 12 for hexagons it gives 9.
There is other way, computing a and h knowing that h is a*sqrt(3) (or going through Pythagorean theorem to get that), and then do computations using units, but it seemed to complicated to do it in 2 minutes.
You can do it! How exciting!
Here I am calculating the altitude of an equilateral triangle of area 1...
R1=3/2
R2=4/2
R3=5/2
Okay, since the triangles are similar by inspection of the angles, the inradii must be similar by same ratio (abc~bde~cef, where f=a+d), so
a=3/4*b
b=3/4*d
a=3/5*c
b=3/5*e
b=4/5*c
d=4/5*e
Then
a=3/4*b=3/5*c=9/16*d=9/20*e
By inradius definition,
r1=TA1/s1=(ab/2)/((a+b+c)/2)
3/2=(a*4/3*a/2)/((a+4/3*a+5/3*a)/2)=a/3
So
a=9/2
b=6
c=15/2
d=8
e=10
And Area=ce=75
The three triangles are similar. Their linear ratios are 3 : 4 : 5
Their hypotenuses correspond to the diagonal, the length and the width of the rectangle. So all three triangles are 3 - 4 - 5 triangles.
We can write the length and width and the diagonal of the rectangle as 4x & 3x & 5x, which happen to be the three sides of the large triangle (the one with the red circle inside), we can write
4x - 5/2 + 3x - 5/2 = 5x
2x = 5, x = 5/2
So the rectangle area = (3x)(4x) = 12x^2 = 12(5/2)^2 = 75
Everything is super clear what you said, except the first equation. Where's that come from?
Ah, okay! I can see it know. It definitely needs way more explanation though.
@ If you focus on the large triangle, its three sides are 3x (rectangle width), 4x (rectangle length) and 5x (rectangle diagonal), the three sides are all tangent lines of the red circle. The 4x is divided into 2.5 and 4x - 2.5 at the point of tangency (the point of contact between the tangent and the circle). The 3x is divided into 2.5 and 3x - 2.5, the 5x is divided into 4x - 2.5 and 3x - 2.5. Therefore 5x = (4x - 2.5) + (3x - 2.5)
@@cyruschang1904 Yes. And all tangent lines of a circle from the same point have equal length. This is the insight I missed.
@@Epyxoid Yes, if two lines that are tangent to a circle intersect at a single point, the point is called the "external point of tangency". The distance between the "external point of tangency" and the point of tangency (where the tangent touches the circle circumference) is the same for the two tangents. It is easy to demonstrate, just connect the circle center to the "external point of tangency“, and to the two points of tangency. you now have two identical right triangles that share the hypotenuse, and they both have the radius as one of their side length.
In January you should do a video with stats for which methods saw the most usage on this month’s puzzles. TSN triangles have been beyond clutch!!
I solved it similarly but using just the side length S, it'd take a very long, but straightforward, proof to show all that even works, but I think showing that all those equilateral triangles fit in that exact shape would be even longer and more tedious work, then again, the equilateral triangle approach gives much faster end calculations so maybe we end up with the same proof length.
Such a beautiful solution. Not going for the trigonometric quickdraw lol
We are gonna hit it!!
How. Exciting.
Dang, that one was easy!
I actually calculated the height and width of the triangles to calculate the area of the rectangle 😂
Same here. I didn't see enough "rigor" in assuming that the rest of the rectangle had a bunch of equilatoral triangles - it kind of "jumped" to that - but knew we could get the height and width of the rectangle from the triangles themselves.
@@Insightfill You know regular hexagons can be tiled with no gaps, yes? That being the case, all you have to do is look at where the sides and vertices of the two hexagons touch the rectangle and it becomes obvious.
@@keith6706 Oh, I get it. It's just that for a channel that usually articulates every small step, it's rare to see a jump like that thrown in.
So the area of each hexagon is 6 and the area of the shaded area is 9.....I SEE what you did there!! 😜😜😜
I feel like we were supposed to find the height and width of the rectangle using the hexagons and find the area with rectangle minus 12
1:50 Thank you!
I was literally looking up equations for trapezoids and trying to find the area of the rectangle, DAMNIT !
Pythagoras got the day off today.
I was hoping for an elaborate trigonometric solution.
My goodness, what a fun and surprisingly short method
Clue for next one: in a right triangle, the incircle has a diameter equal to (short side 1 + Short side 2 - hypotenuse)
literally solved it as you uploaded. lets go.
My one is coming 12
How would you do this algebraically?
Not often I can get one just looking at the thumbnail on my phone
Ooh the next one's easy if you know that a 3-4-5 triangle has an incircle with radius 1. The three triangles must be scaled 3-4-5 so a 3-4-5 triangle is what will work. So the largest triangle has legs 15 and 20, those are the sides of the rectangle, so the rectangle's area is 300 sq u.
close, but the diameter is 5, which means the radius is 5/2. so the largest triangle actually has legs 15/2 and 20/2 which means the area is 75
@@leolin3207 Oh crap that's the diameter. Yeah, so since the radius is half of what I thought, the area is a quarter of what I thought since (1/2)^2 = 1/4. 75 sq u, final answer Regis.
Your first issue: how do you know those are 3-4-5 scaled triangles?
@@keith6706 Look at how a leg of one plus a leg of another equals the hypotenuse of the third. Since the inradius (or in-diameter in this case) of two similar triangles scales at the same ratio as the sides, the largest triangle has a hypotenuse that's 5/3 the hypotenuse of the smallest one. The math is easy, but since I've seen this configuration a lot I didn't need to scribble it out. I'm sure I;ve mentioned enough to make that path obvious.
You were clean shaven on day 21 puzzle, had 2 to 3 days growth of beard on day 22 puzzle and are clean shaven on day 23. Let’s use a 30 60 90 triangle to see how often Andy shaves his face. 😂😂
what tool you use, is it just Google slides or PowerPoint or an math based program.
Is it 8? Not mathy but just make everything into little half triangles.
Why does it want to know the answer?
Let's assume that the length of the rectangle is a, its width is b, and its diameter is c. We get the equations a+b-c=5, a²+ab-ac=4c, b²+ab-bc=3c. By solving this set of equations, we find a=10, b=15/2, c=25/2. So the area of the rectangle is 10*15/2=75.
Wait, it can't be just counting, can it??
I love ur vids
I'm almost certain you are going to manage to do this in a much simpler way than I did.
edit: oh that was clever.
Found the answer for tomorrow's puzzle.
I knew it was gonna be a sea of equations and tons of algebra work, just didn't know whether or not I'd be able to get every single length, thought maybe I'd just get a L*W = sth, at the end.
Anyways, SPOILER FOR THE SOLUTION.
The answer I got for the area of the rectangle was 75u².
Ok, in order to get that I needed to create from scratch around 5 main equations for 5 unknowns.
Now obviously I had a lot of extra equations just in case (only 5 were needed to get my 5 unknowns, not sure which were unnecessary), and obviously I'm gonna write the actual equations here, I'm only gonna write their type, how I got each, the main idea, and how many I got of each.
First off, there's a formula for circles inside right triangles (Incircles) that has the inradius in it (radius of the incircle).
R = (a + b - c)/2, where a, b, and c, are the sides and c is the hypotenuse.
Made three of those for each incircle.
We need 2 more equations.
I avoided using the equations with Pythagoras theorem, since they'd get me squares and square roots and make the equations much more complicated, so I used similarities.
My variables were: the rectangle length, and width, the side from the upper corner to the diagonal, and the 2 pieces it was separated into (which makes 5 unknowns), named: L, W, b, a, and c.
Each of those triangle is similar to the 2 others via AA (NOT CONGRUENT THOUGH).
2 similarity equations are more than enough.
Now getting the exact length of b was a piece of cake (the common leg of the smallest triangles), I just used the first 3 equations, addion and substitution with the third, and boom, got b = 6 (just by eye observation).
Getting the other 4 was the part that needed a lot of work.
Some editing to the 5 equations after getting the b, then some algebra got me that, L = 10, W = 7.5
I noticed that you wrote a lot, brother, the issue is not that difficult, and at first, it appears to you that the equations are complicated, and this is not true. To clarify this, let us assume that the length of the rectangle is a, its width is b, and its diameter is c. In each right triangle, we write the equation that connects the sides of the right triangle with the radius of the circle drawn inside it, so we get the following equations: a+b-c=5, a²+ab-ac=4c, b²+ab-bc=3c. We substitute in the second equation b-c=5-a, and substitute in the third equation a-c=5-b, so we get 5a=4c, 5b=3c. Here we get an equation with one unknown c, which is 4c/5+3c/5-c=5, and from it c=25/2, so a=10 and b=15/2, so the area of the rectangle is equal to 10*15/2=75. You can notice from the beginning that the sides of the right triangle are proportional to the diameters of the circles, i.e. a=4x, b=3x, and c=5x, where there is one unknown we are looking for and we find it x=5/2.
Well. That was easy
Isn't that a lot of assuming though? When reading blueprints we are taught never to assume just because of looks.
My question as well.
No, because it says that is a regular hexagon
There's no assumption. It's insight acquired by experience. We can see that the two hexagons touch the rectangle with two sides and two corners, which means what he did is completely works.
if you know what your doing its super simple. if not.... yea good luck. LOL
Day 24: 75 sq units
This isnt a mathematical solution. What would be one if done mathematically ?
It is, you can prove it all geometrically by using complementary angles and similar.
@@KrytenKoro That is what I was expecting from this video.
@@rdgspams5669
Geometry is a part of mathematics.
Since it is a regular hexagon meaning all sides and angles are the same, using the formula for sum of angles in a convex polygon (n-2)*180° based on number, n, of sides, each angle will be (6-2)*180°/6=120°. Any line segments constructed within a regular polygon from one of its vertices that creates two halves, i.e. equal area polygons, must be angle bisectors at the vertex, and the midpoint of this line segment is the center of the regular polygon, so drawing three such line segments within the regular hexagon will create six triangles with two 60° angles, therefore the third angle must also be 60°, since the sum of angles in a triangle is (3-2)*180°=180°. The "tiling" of the rest of the rectangle, follows from using equilateral triangles with area of 1. All the triangles with area 1/2 occur because the rectangle boundaries cut the equilateral triangle in half, which, since an equilateral triangle is a regular polygon, then as explained previously means by definition of an angle bisector or (1/2)*60°=30° at the vertex and one 60° angle still exists, so the other must be 90° (which can easily be verified at two of the corners of the rectangle).
Geometry is a part of mathematics.
Since it is a regular hexagon, by definition, all sides are of equal length and all angles are equal in measure. Using the formula for sum of angles in a convex polygon (n-2)*180° based on the number, n, of sides, each angle will be (6-2)*180°/6=120°. Any line segments constructed within a regular polygon from one of its vertices that creates congruent halves of the polygon must be angle bisectors at the vertex, and the midpoints of these line segments coincide since they would be diameters of a circumscribed circle, i.e. a circle upon which all vertices of the polygon lie, so constructing three such line segments within the regular hexagon will create six triangles with two 60° angles, therefore the third angle must also be 60°, since the sum of angles in a triangle is (3-2)*180°=180°. The "tiling" of the rest of the rectangle follows placing equilateral triangles with area of 1 side by side, just as they appear in the hexagon. Since an equilateral triangle is a regular polygon, then the triangles with area 1/2 occur because the rectangle boundaries cut any overlapping equilateral triangles with area 1 in half
@@bryanalexander1839 Thanks Bryan
Was the answer 8?
Ohhh no it's 9
You shaved!
W
*H O W*
*E X C I T I N G*
والله يا انا طولتها طلعت معي في الأخير 9.948222 أكيد في شيء فاتني في الحساب