Andy! You need to start a merch store. I would totally buy a t shirt with one of your catchphrases on it. "Let's put a box around it.", "How exciting.", "I'm gonna solve it in 3, 2, 1."
Hey guys, I made this T-shirt and I put a box around it. I’m going to wear it in 3, 2, 1. Ok first I need to take off the shirt I’m wearing. How exciting.
I wish school taught me to find solutions without the unnecessary calculations. He found the area without substituting for an area formula, just using formulas as is
It's genuinely comforting watching him do his thing with a polite voice and a quiet smile on his face. I never thought I'd have a "comfort UA-camr" but he is mine
Here's a fun trick you can do to solve this almost instantly. Since no lengths are specified, we can assume that the figure is drawn such that the width of the rectangle is equal to its height and therefore is also a square. Since the both squares have midpoint on the center and 2 opposite corners on the circle, they must be the same area. (Try this on desmos or another geometry tool) From there, you can flip one half of the kite along the long axis. since it's symmetric and right, that means the shape will be a rectangle with one side being the length of the square and the other half that. Therefore, the kite shape is half the size of either square, so each must be 2×15. Both full squares together would then be 60. We already know the overlap is 15, so we subtract 15 for the to avoid double counting, and then another 15 because the area is purple.
My solution: I labelled the side of the square "s", the height of the rectangle "h", and divided the width of the rectangle into two equal parts w. So the area of the square is s^2, and the rectangle is 2wh. So the equation for the solution is A = 2wh + s^2 - 30 I divided the pink area into two parts (an and b) as Andy did, and recognized that a is 1/4 the area of the square. The area of triangle b is then 1/2*w*h. We know the sum of these two is 15. Putting this together we get: 1/2 * w * h + ( s^2 / 4 ) = 15 Multiply both sides of this equation by 4 and you get 2wh + s^2 = 60 You can substitute that into my first equation and get A = 60 - 30 And that gives the answer 30
For tomorrow: 9u². You can make a full hexagon from the chunks above and below the labeled ones. Then, you can rearrange the remaining pieces (with proper justification, of course) into another half hexagon.
You can't determine the lengths as there are infinitely many variations that you could draw to fit this diagram by sliding the top point of the square along the circle while adjusting radius accordingly.
For the next problem, you can partition the orange area into equilateral triangles that have been split in half. A regular hexagon is made of 6 equilateral triangles, so the area of each one is 1, so the split pieces have area 1/2. Count them up: 18 such pieces make the orange area. So the area is 18/2 = 9.
Ok, that was a nice way of finding the area. It could've been done much faster using a bit less geometry and almost the same amount of algebra work. After finding the area a and area b congruence, we could just say that the 15 area can be separated into 2 parts by drawing a radius through it to the point in it on the circumference. That divides it into 2 right triangles, if we label the rectangle dimensions L and W (for length and width), and the square side length S, we get that the left triangle has legs L/2 and W, and the upper triangle has legs S/2 and S. From there we can calculate the areas of the 2 and equate them to 15, so, 1/2 * L/2 * W + 1/2 * S/2 * S = 15, L*W/4 + S²/4 = 15. We're still looking for A = LW + S² -30, so now we have, LW/4 + S²/4 = 15 (multiply by 4), hence, LW + S² = 60, substituting in the A equation, A = LW + S² - 30, A = 60 - 30 = 30
I did this. It only took me 3-4 hours, and my solution looks nothing like this! I always think analytical geometry and coordinates, and information versus independent unknowns. It can come out really messy at times, but it worked nicely for this. It only took me so long because I am so error prone. I am the "rip it up, start again" man.
I set a as side length of square, b as shorter length of rectangle, c as longer length of rectangle. Area of yellow = a² + bc - 30 I know you can join 2 radii to the 2 corners of the square making isosceles triangle. Half it and you get right angle triangle with side lengths ½a, a and r where r is radius. I use Pythagoras theorem to get value for a² which is ⅘r² Now area of yellow = ⅘r² + bc - 30 Then i work out area of when you connect radius to top corner of square splitting the pink area into 2 right triangles. I choose the one where lengths of the triangle is b, ½c and r and i get the area of that triangle as bc/4 then i find area of the other pink triangle which i know has lengths of ½a, a and r. Area is a²/4 and i knew a² was ⅘r² so i substituted that in and added the 2 pink areas together = 15 So bc/4 + ⅕r² = 15 i multiply everything by 4 to get bc + ⅘r² = 60 The area of the yellow was ⅘r² + bc - 30 Substitute 60 for ⅘r²+bc To get 60 - 30 = area of yellow 60 - 30 = 30 u²
Wow I did this in a completely different way, actually working out the lengths of the sides of the square and rectangle in terms of 's' (side of the square), where s^2 worked out to be 300/11, the side of the rectangle is 2sroot5 / 5, half the base of the rectangle is 3sroot5 / 10, etc. Your solution is of course far more elegant.
THe tomorrow puzzle: I started by trying to figure out the side of the hexagon using A = 3sqrt(3)s^2/2, but that was a mess. So here's what I did instead. I saw the little 30-60-90 triangles in the corner. We can make one hexagon into 12 of them, so the area of the little triangle is 1/2. Then you just count all the triangles.
Answer to the next question: A hexagon of area 6 is composed of 6 equilateral triangles of area 1 with side length 2√(1/√3) and height √(√3) Rectangle length = 7√(1/√3) Rectangle width = 3√(√3) Red area = 21 - 12 = 9
I expected that the area of the yellow region would be twice the area of the red region, but I calculated the length of the side of the square and found that a=(10√33/11). Then I calculated the area of the yellow region and found that it was equal to (11a²/5)-30=30.
I also did tomorrow's question and got the same result. My first attempt lead me down an algebra rabbit hole. Then I took a step back and realized that if each hexagon contains 6 equilateral triangles with the area of 1, then the rectangle is comprised of 3 rows of 7 equilateral triangles . The total area of the rectangle is 21, and you subtract 12, leaving 9 units squared.
Isn't Yellow = x+y-15? There's only one pink area, why do we subtract it twice. i get we got a rectangle and a square, but they intersect, so it's still only once that we have to subtract it
failed yesterday's. am deeply shamed at how simple you made it look tomorrow is 9 add in lines from the empty corners to the closest hexagon corners to get the bottom half of a hexagon the top half of a hexagon (one hexagon) two corner triangles each a twelfth of a hexagon and two remaining triangles, both a sixth of a hexagon which makes 1.5 hexagons in the shaded area, which is 9
That seems like a distinction without a difference. Or in the spirit of the video, those seem like congruent terms (square units and units²). Like if the units were meters, you could say "square meters" or "meters square." And you'd probably write "m²" no matter how you say it.
so... considering both the areas of the rectangle and the square can be divided in 4 equal parts labeled as "a" and "b" you can just consider their overlap and simplify the yellow area directly to 2a + 2b = 2 (15) or 30 units²
Andy! You need to start a merch store. I would totally buy a t shirt with one of your catchphrases on it. "Let's put a box around it.", "How exciting.", "I'm gonna solve it in 3, 2, 1."
Hey guys, I made this T-shirt and I put a box around it. I’m going to wear it in 3, 2, 1. Ok first I need to take off the shirt I’m wearing. How exciting.
@@TimMaddux beautifull
Bro thought no one would find the reason for naming that ∆NDY 1:50
(His name)
HOW EXCITING.....!
Wow! Good find!
On the other side of the equation he wrote ΔSKY, which could be read as "ASK Y" or "ask why". Coincidence? I'd like to think not. How exciting.✌🏻🤓
I wish school taught me to find solutions without the unnecessary calculations. He found the area without substituting for an area formula, just using formulas as is
Andy Sky
My bro is living in his own wonderful quiet beautiful world. ❤
It's genuinely comforting watching him do his thing with a polite voice and a quiet smile on his face.
I never thought I'd have a "comfort UA-camr" but he is mine
@@mr.x991yeah man❤.
In his own exciting world
For the hexagons, by inspection the orange area is 1.5 hexagons, so the shaded area is 9
Every solution is more exciting than the one the day before.
Always feel exciting seeing you solve the puzzle. Thank you.
I never would have found this one
Here's a fun trick you can do to solve this almost instantly.
Since no lengths are specified, we can assume that the figure is drawn such that the width of the rectangle is equal to its height and therefore is also a square. Since the both squares have midpoint on the center and 2 opposite corners on the circle, they must be the same area. (Try this on desmos or another geometry tool)
From there, you can flip one half of the kite along the long axis. since it's symmetric and right, that means the shape will be a rectangle with one side being the length of the square and the other half that. Therefore, the kite shape is half the size of either square, so each must be 2×15.
Both full squares together would then be 60. We already know the overlap is 15, so we subtract 15 for the to avoid double counting, and then another 15 because the area is purple.
My solution: I labelled the side of the square "s", the height of the rectangle "h", and divided the width of the rectangle into two equal parts w. So the area of the square is s^2, and the rectangle is 2wh. So the equation for the solution is A = 2wh + s^2 - 30
I divided the pink area into two parts (an and b) as Andy did, and recognized that a is 1/4 the area of the square. The area of triangle b is then 1/2*w*h. We know the sum of these two is 15. Putting this together we get:
1/2 * w * h + ( s^2 / 4 ) = 15
Multiply both sides of this equation by 4 and you get
2wh + s^2 = 60
You can substitute that into my first equation and get A = 60 - 30
And that gives the answer 30
9 is area of the shaded region
For tomorrow: 9u².
You can make a full hexagon from the chunks above and below the labeled ones.
Then, you can rearrange the remaining pieces (with proper justification, of course) into another half hexagon.
The side of the square = 10sqrt(3/11), the sides of the rectangle are 4sqrt(15/11) and 6sqrt(15/11)
You can't determine the lengths as there are infinitely many variations that you could draw to fit this diagram by sliding the top point of the square along the circle while adjusting radius accordingly.
I know. Exciting, huh?
For the next problem, you can partition the orange area into equilateral triangles that have been split in half. A regular hexagon is made of 6 equilateral triangles, so the area of each one is 1, so the split pieces have area 1/2. Count them up: 18 such pieces make the orange area. So the area is 18/2 = 9.
Casual geometry. Very nice.
I'm so addicted with your explanation. So simple and understandable even for D-math student like me 😭
Very nice, interesting and intricate math geometry puzzle as always.👏👏👏
Ok, that was a nice way of finding the area.
It could've been done much faster using a bit less geometry and almost the same amount of algebra work.
After finding the area a and area b congruence, we could just say that the 15 area can be separated into 2 parts by drawing a radius through it to the point in it on the circumference.
That divides it into 2 right triangles, if we label the rectangle dimensions L and W (for length and width), and the square side length S, we get that the left triangle has legs L/2 and W, and the upper triangle has legs S/2 and S.
From there we can calculate the areas of the 2 and equate them to 15, so, 1/2 * L/2 * W + 1/2 * S/2 * S = 15, L*W/4 + S²/4 = 15.
We're still looking for A = LW + S² -30, so now we have, LW/4 + S²/4 = 15 (multiply by 4), hence, LW + S² = 60, substituting in the A equation, A = LW + S² - 30, A = 60 - 30 = 30
I did this. It only took me 3-4 hours, and my solution looks nothing like this!
I always think analytical geometry and coordinates, and information versus independent unknowns. It can come out really messy at times, but it worked nicely for this. It only took me so long because I am so error prone. I am the "rip it up, start again" man.
Respect sir ❤❤❤❤You make mathematics enjoyable for us.
This videos are so interesting, thanks for sharing!
Andy sky!
I set a as side length of square, b as shorter length of rectangle, c as longer length of rectangle.
Area of yellow = a² + bc - 30
I know you can join 2 radii to the 2 corners of the square making isosceles triangle. Half it and you get right angle triangle with side lengths ½a, a and r where r is radius. I use Pythagoras theorem to get value for a² which is ⅘r²
Now area of yellow = ⅘r² + bc - 30
Then i work out area of when you connect radius to top corner of square splitting the pink area into 2 right triangles. I choose the one where lengths of the triangle is b, ½c and r and i get the area of that triangle as bc/4 then i find area of the other pink triangle which i know has lengths of ½a, a and r. Area is a²/4 and i knew a² was ⅘r² so i substituted that in and added the 2 pink areas together = 15
So bc/4 + ⅕r² = 15 i multiply everything by 4 to get bc + ⅘r² = 60
The area of the yellow was ⅘r² + bc - 30
Substitute 60 for ⅘r²+bc
To get 60 - 30 = area of yellow
60 - 30 = 30 u²
Rare moment of not using 30 - 60 - 90 triangle
It's all about those right triangles.
This one was extra interesting to me for some reason. Cool answer
Wow I did this in a completely different way, actually working out the lengths of the sides of the square and rectangle in terms of 's' (side of the square), where s^2 worked out to be 300/11, the side of the rectangle is 2sroot5 / 5, half the base of the rectangle is 3sroot5 / 10, etc. Your solution is of course far more elegant.
THe tomorrow puzzle: I started by trying to figure out the side of the hexagon using A = 3sqrt(3)s^2/2, but that was a mess. So here's what I did instead. I saw the little 30-60-90 triangles in the corner. We can make one hexagon into 12 of them, so the area of the little triangle is 1/2. Then you just count all the triangles.
Answer to the next question:
A hexagon of area 6 is composed of 6 equilateral triangles of area 1 with side length 2√(1/√3) and height √(√3)
Rectangle length = 7√(1/√3)
Rectangle width = 3√(√3)
Red area = 21 - 12 = 9
He's starting to catch up now
He’s beginning to believe
I expected that the area of the yellow region would be twice the area of the red region, but I calculated the length of the side of the square and found that a=(10√33/11). Then I calculated the area of the yellow region and found that it was equal to (11a²/5)-30=30.
👏👏👏👏👏👏
0:07 how are we gonna catch up
Answer is 9, I split the hexagons into triangles of area 1.
❤
First it looked more difficult than it was.
Area shades = 9 units squared ????
I also did tomorrow's question and got the same result. My first attempt lead me down an algebra rabbit hole. Then I took a step back and realized that if each hexagon contains 6 equilateral triangles with the area of 1, then the rectangle is comprised of 3 rows of 7 equilateral triangles . The total area of the rectangle is 21, and you subtract 12, leaving 9 units squared.
@@benjaminmichael5719 yeap, it's easy to visualise and arrive at 9unit². But one has to go step by step for formal explanation.
Isn't Yellow = x+y-15? There's only one pink area, why do we subtract it twice. i get we got a rectangle and a square, but they intersect, so it's still only once that we have to subtract it
That was so coool
this one was very hard
Day 23: 9 sq units
How do you know that the intersection of the square and rectangle is at their centers and not just the circles?
Application of the chord bisector theorem can easily show it to be so.
tri NDY = ANDY 1:50
The triangle ∆NDY got me
How can I learn to solve this kind of problems?
failed yesterday's. am deeply shamed at how simple you made it look
tomorrow is 9
add in lines from the empty corners to the closest hexagon corners to get
the bottom half of a hexagon
the top half of a hexagon (one hexagon)
two corner triangles each a twelfth of a hexagon
and two remaining triangles, both a sixth of a hexagon
which makes 1.5 hexagons in the shaded area, which is 9
day23
6+3=9😊
Are those yellow triangles 30:60:90s? Or am I going mad?
They are, since the little edge is exactly half the long edge
@@coogs5688 sorry I was mistaken - hypotenuse (not long leg) of a 30-60-90 triangle is twice the short leg. These are not 30-60-90 triangles
is the answer 9 for the tomorrow's question?
No 30-60-90 triangle :(
U r Cool
How is it x+y-30 and not x+y-15 ?
When taking the area of both squares individually, the 15 overlaps, so you have to count it twice.
Next problem: 9
How did bro grow a full beard and mustache in the span of 12 hours?
IT’S sq. units NOT u^2 AHHHHHHHHHHHHH
MY EYES ARE BURNING!!!!
That seems like a distinction without a difference. Or in the spirit of the video, those seem like congruent terms (square units and units²). Like if the units were meters, you could say "square meters" or "meters square." And you'd probably write "m²" no matter how you say it.
Hope you're doing ok during the holidays while putting out these videos. 😢
IT’S RHS(RIGHT-ANGLE HYPOTENUSE SIDE) NOT HL THEOREM AHHHHHHHHHHHH
so... considering both the areas of the rectangle and the square can be divided in 4 equal parts labeled as "a" and "b"
you can just consider their overlap and simplify the yellow area directly to 2a + 2b = 2 (15) or 30 units²
My math bad
hi
Is the answer to the next question [6+7sqrt(3)]/3? Which about 6.0415
Unfortunately it is not. The answer is an integer.
Bro shave your beard! It looks “how exiting” btw in 1 min
Hell yeah, 3rd comment!
Next problem: by inspection, 7 sq units
It's all easy for me cause I'm indian(Asian)
Bro?
🤦🏻♂