He got a bit excited because it's basically just the same problem once again. The base of the rectangle equals twice the radius of all three circles i.e. 2 x 1 x 3 = 6. Then we can just jook at one triangle of the rectangle and its inscribed circle (say, the fully cyan one) and use our approach from this problem once again to find the height of the triangle and thereby the height of the rectangle.
I've noticed that a few of these puzzles this month have had assumptions baked into them that aren't explained. It's been a little frustrating because when presented with the initial puzzle, I usually can't start to solve it at all.
Sometimes the assumptions are combined into the problem because otherwise it’s not solvable, but sometimes it’s included in the problem, but he doesn’t include it when he’s pasting the picture into the video.
For those attempting to solve day 31 using trigonometry, it is worth noting that, from the 5-12-13 right triangle, if tan(2Θ) = 5/12, then tan(Θ) = 1/5, and, if tan(2Θ) = 12/5, then tan(Θ) = 2/3.
It's actually sad that I couldn't care less for maths in high school, while discovering the beauty of it 10 years later.. I just wish mathematics felt more like a challenge than a chore back then..
Maybe I missed it, but how do we know that the height arrow (4) is actually tangent to the two circles? I’m not sure that’s a given. I can assume that the circles are tangent to the sides, but I’m not sure the height arrow counts for that since it’s a measurement and not a drawn line per se. Correct me if I’m wrong!
Without that information the problem is impossible to solve as the circles could then be arbitrarily far apart from one another, while still making a 4-high triangle. Which makes the base, and the area, unbounded. We wouldn't even be able to say that 21 is the lower bound on the area, as we could move the circles closer together before the touch eachother. So either the problem has no solution or the height is tangent to both circles, and it becomes practical to evaluate what the intent of the question is.
Let's look at the x tangent lines. The distance from the intersecting points that the x's make to the tangent of the circle is x, but the distance from that point to the other intersecting point (the peak of the triangle) is what I'll try to explain to you as well as explain how it's tangent. So let's call that other distance z (not to confuse it with the already written y's). Now to look at the bottom. The distance from the intersecting distance to the tangent is x, and we've proven that from the tangent to the edge of the circle we get the distance of 1,5. We'll draw a vertical line perpendicular to the end of the circle, the 1,5 distance (so at a right angle) and when we keep going up until we get to the point of intersection with z, you'll find that that point of intersection is indeed the peak of the triangle. And such distance from the base to the top is the height of the triangle itself. It might seem confusing at first, but imagine the two orange lines drawn from step one, since those two distances are equal and are both tangent, that means that the one going downward is perpendicular to the horizontal radius, and therefore it'll be at a right angle, meaning it's perpendicular to the base of which we measure the distance of 1,5.
Let us assume that x is the width of the rectangle, then the diagonal of this rectangle is equal to x + 4. According to the Pythagorean theorem, (x + 4)² = 6² + x², so x = 5/2. Therefore, the area of the rectangle is equal to 6 * 5/2 = 15.
The drawing didn't make it too obvious that the 2 circles are touching, personally, I thought they weren't which made it impossible to solve for me. Anyways, great video l, and your solving technique was awesome, thanks again for a job well done
@@Grizzly01-vr4pn that's exactly what I thought at the start too, but he solved it as the 2 circles touching, when he added the base distances and made them ewual to the whole base. If the circles weren't touching there'd be a difference in distance, equal to the width of the altitude, but in this case there wasn't
@@Z-eng0 Not so. If you follow the video, it should become apparent that if the circles _were_ touching, the method he uses would not work. Also, line segments are assumed to have zero width.
@Grizzly01-vr4pn I did follow the video, but ok, bear with me for a minute. The first few steps he did, didn't need to have touching circles, it works in both cases, from the moment he assumed the distance 4 - 1.5 = 3, is the same as the left sided minus x, that was an assumption that the altitude drawn is touching the left circle (since it's tangent to 🔴 r=3). And he did the same for the 🔴 r=2. And since we know all lines have width zero, and it's touching bith circles, that only leaves the option that the 2 circles are also touching. 1 line touching 2 circles at the same time means the 2 circles are touching.
using (a+b-c)/2 to get r . 1rst triangle 4+b-c=3, mean c-b =1. pythagore 4x4 + bxb = (b+1)(b+1) ... same for the other triangle wonder wich is more simple. :)
15 take the isosceles triangle between the bottom left corner to the tangents of the right hand circle the bottom leg is five making the top leg 5 plugging those figures into the right angled triangle bottom right bottom side is 6, right side is 1+x and hypotenuse is 5+x pythag says 6^2+(1+x)^2=(5+x)^2 36+1+2x+x^2=25+10x+x^2 36+1-25=x^2-x^2+10x-2X 12=8x x=1.5 making the side 2.5 area=6*2.5=15
Excellent, I got that too! In these Aggvents, many tangents were assumed. It’s important to assume the circles are tangent to the vertical lines, and Not to each other.
Can I add a line that is the diameter of the larger circle (length : 3cm) and use the similar triangle (AAA) to find the base length of the triangle is x+1.5+2 3÷4=x÷(x+1.5) (corr. sides , ~= triangle) x=4.5 So the area of the triangle is (x+1.5+2)×4÷2 = 32÷2 = 16 cm² Is it correct ?
Radius of the bigger circle, which has a diameter of 3, so half of that. And the radius in this case is the distance between the tangent point "on the bottom" of the circle and the light blue tangent line....
The next and the last aggvent calendar problem: If the rectangle width = 2 + a, the diagonal = 6√(a^2 + 1). So we can write √[(2 + a)^2 + 6^2] = 6√(a^2 + 1) (2 + a)^2 + 6^2 = 36(a^2 + 1) 35a^2 - 4a - 4 = 0 (7a + 2)(5a - 2) = 0 a = 2/5 Rectangle area = 6(2 + a) = 12 + 12/5 = 14.4
If x+2 is the width of the rectangle and 6 is its length, then the diagonal of the rectangle is the hypotenuse of the right triangle that represents half of the rectangle, and the circle drawn inside this triangle has a radius of 1, so the diagonal is equal to x+2+6-2r=x+2+6-2=x+6, so the right triangle has sides x+2,6,x+6, then we write the Pythagorean theorem to determine x.
I'm stumped on tomorrow's! EDIT: Never mind! I found a fairly simple solution using Pythagorean theorem and the rule of intersecting tangents: Let x be the height of the rectangle: 6^2 + x^2 = (5+(x-1))^2... x = 5/2. So rectangle is exactly 15 units squared. :)
It's similar to current. Base is x + 1, height is y + 1, hypotenuse is x + y. Because of circles, base is 6, which gives x == 5, then Pythagorean theorem kicks in and gives y == 1.5, and then 6 x 2.5 == 15.
Use intersecting tangents to define a variable that is the same distance from tangent to either yellow or blue circle to one corner of one triangle, then use Pythagoras to solve for that distance. This will give you the height of the triangle thus height of the rectangle thus the area. And it's not 12sqrt3
the base of the big rectangle is 6*(the radius) = 6 , using the intersecting teangents : the diagonal is X+Y but X=6-1 =5 so the diagonal is 5+Y , the height is Y+1 using the pythagore theorem : 6² + (1 + Y)² = (5 + Y)² Y² + 2Y + 37 = Y² + 10x + 25 Y = 1.5 so the height is 1+1,5= 2,5 the area is 2,5 * 6= 15 (sorry if english is bad , not my language)
As an Indian who dreams of coming to the United States to contribute to its diversity, I deeply admire the values of freedom and opportunity that this nation represents. However, I’d like to understand how native Americans view immigrants and the challenges of racism. Do you think America is welcoming to immigrants, or does racism still pose a significant barrier for those seeking to build a life here?"
"Does that make sense y?"
Peak comedy
Andy comedy....
Absolute Cinema
Never heard Andy laugh before. My day is complete.
Those notes came in clutch multiple times on this one; this has a really clean solution too - nice!
The jump scare at the end
He got a bit excited because it's basically just the same problem once again. The base of the rectangle equals twice the radius of all three circles i.e. 2 x 1 x 3 = 6. Then we can just jook at one triangle of the rectangle and its inscribed circle (say, the fully cyan one) and use our approach from this problem once again to find the height of the triangle and thereby the height of the rectangle.
How. Exciting.
I love how it’s 2:30 in the morning for me rn😂nothin like solving math problems while being sleep deprived 🔥
5:30 for me, we’re in the same boat lol
Im watching this at 2:30am as well lol, what are the odds
Keep going with your content andy! It is very educative
Only one more aggvent puzzle to go Andy. How exciting! 😊
a=a. Once you get that, the rest is easy! LOL!
Why did you assume the light blue line to be tangent to the circles? If the circles are further apart, the calculation doesn't work out.
I don't think it's solvable without making that assumption
I've noticed that a few of these puzzles this month have had assumptions baked into them that aren't explained. It's been a little frustrating because when presented with the initial puzzle, I usually can't start to solve it at all.
Sometimes the assumptions are combined into the problem because otherwise it’s not solvable, but sometimes it’s included in the problem, but he doesn’t include it when he’s pasting the picture into the video.
It’s by design and as drawn. If not, he would have to specify the distance between the blue line and the two circles.
All the same, I think that it should be clarified in the problem
Hey, I did the last one entirely on my own! (It helps that it uses pretty much exactly the same concept as this one.)
“Such brilliant notes. Speaking of brilliant, let’s talk about Brilliant.”
00:41 that laugh
1 more to gooo!
For those attempting to solve day 31 using trigonometry, it is worth noting that, from the 5-12-13 right triangle, if tan(2Θ) = 5/12, then tan(Θ) = 1/5, and, if tan(2Θ) = 12/5, then tan(Θ) = 2/3.
It's actually sad that I couldn't care less for maths in high school, while discovering the beauty of it 10 years later.. I just wish mathematics felt more like a challenge than a chore back then..
How do you know that the visualized high of the triangle is actually the tangent line for both of the circles? Is it just from the picture?
I've really been enjoying these
the problem really doesn't make it clear if the circles are tangent to the altitude or not
Maybe I missed it, but how do we know that the height arrow (4) is actually tangent to the two circles? I’m not sure that’s a given.
I can assume that the circles are tangent to the sides, but I’m not sure the height arrow counts for that since it’s a measurement and not a drawn line per se.
Correct me if I’m wrong!
Without that information the problem is impossible to solve as the circles could then be arbitrarily far apart from one another, while still making a 4-high triangle. Which makes the base, and the area, unbounded. We wouldn't even be able to say that 21 is the lower bound on the area, as we could move the circles closer together before the touch eachother.
So either the problem has no solution or the height is tangent to both circles, and it becomes practical to evaluate what the intent of the question is.
Let's look at the x tangent lines.
The distance from the intersecting points that the x's make to the tangent of the circle is x, but the distance from that point to the other intersecting point (the peak of the triangle) is what I'll try to explain to you as well as explain how it's tangent. So let's call that other distance z (not to confuse it with the already written y's).
Now to look at the bottom. The distance from the intersecting distance to the tangent is x, and we've proven that from the tangent to the edge of the circle we get the distance of 1,5.
We'll draw a vertical line perpendicular to the end of the circle, the 1,5 distance (so at a right angle) and when we keep going up until we get to the point of intersection with z, you'll find that that point of intersection is indeed the peak of the triangle. And such distance from the base to the top is the height of the triangle itself.
It might seem confusing at first, but imagine the two orange lines drawn from step one, since those two distances are equal and are both tangent, that means that the one going downward is perpendicular to the horizontal radius, and therefore it'll be at a right angle, meaning it's perpendicular to the base of which we measure the distance of 1,5.
@@foolishgold3171 Thank you!!
@@phiefer3 Thank you!
Starting off the new year with triangles.
How exciting!
It’s still December. How exciting!
Let us assume that x is the width of the rectangle, then the diagonal of this rectangle is equal to x + 4. According to the Pythagorean theorem, (x + 4)² = 6² + x², so x = 5/2. Therefore, the area of the rectangle is equal to 6 * 5/2 = 15.
Its a trangle if you know the hight and the base width you can just multiply and half, you would have been done by the two minute mark.
The drawing didn't make it too obvious that the 2 circles are touching, personally, I thought they weren't which made it impossible to solve for me.
Anyways, great video l, and your solving technique was awesome, thanks again for a job well done
They aren't touching
Each circle is tangent to the altitude (line labelled '4') but they aren't tangent to each other.
@@Grizzly01-vr4pn that's exactly what I thought at the start too, but he solved it as the 2 circles touching, when he added the base distances and made them ewual to the whole base.
If the circles weren't touching there'd be a difference in distance, equal to the width of the altitude, but in this case there wasn't
@@Z-eng0 Not so.
If you follow the video, it should become apparent that if the circles _were_ touching, the method he uses would not work.
Also, line segments are assumed to have zero width.
@Grizzly01-vr4pn I did follow the video, but ok, bear with me for a minute.
The first few steps he did, didn't need to have touching circles, it works in both cases, from the moment he assumed the distance 4 - 1.5 = 3, is the same as the left sided minus x, that was an assumption that the altitude drawn is touching the left circle (since it's tangent to 🔴 r=3).
And he did the same for the 🔴 r=2.
And since we know all lines have width zero, and it's touching bith circles, that only leaves the option that the 2 circles are also touching.
1 line touching 2 circles at the same time means the 2 circles are touching.
After done all aggvent calendar could u still make video at least per 2 days? We love to see ur video ❤
first time saw Andy laugh
I put into Desmos (y-r)^2+(x-r)^2=(r)^2 and y=mx+b, then replaced r with 2, then b with 4, and messed with m until it touched the circle, and yeah.
6² + (1 + x)² = (5 + x)²
x² + 2x + 37 = x² + 10x + 25
x = 1.5
area = 6 × 2.5 = 15
Can we have an Aggvent for January, please?
that was a really nice solve.
using (a+b-c)/2 to get r . 1rst triangle 4+b-c=3, mean c-b =1. pythagore 4x4 + bxb = (b+1)(b+1) ... same for the other triangle wonder wich is more simple. :)
15
take the isosceles triangle between the bottom left corner to the tangents of the right hand circle
the bottom leg is five making the top leg 5
plugging those figures into the right angled triangle bottom right
bottom side is 6, right side is 1+x and hypotenuse is 5+x
pythag says 6^2+(1+x)^2=(5+x)^2
36+1+2x+x^2=25+10x+x^2
36+1-25=x^2-x^2+10x-2X
12=8x
x=1.5
making the side 2.5
area=6*2.5=15
Excellent, I got that too!
In these Aggvents, many tangents were assumed. It’s important to assume the circles are tangent to the vertical lines, and Not to each other.
mission failed... the new year has already started
Awesome work.
One more to go. This was exciting!
Can I add a line that is the diameter of the larger circle (length : 3cm) and use the similar triangle (AAA)
to find the base length of the triangle is x+1.5+2
3÷4=x÷(x+1.5) (corr. sides , ~= triangle)
x=4.5
So the area of the triangle is (x+1.5+2)×4÷2 = 32÷2 = 16 cm²
Is it correct ?
Can't you solve this by taking out two triangles at the top one with base 1,5 and height 1 and then 1/1,5 = 4 / x and tdo the same for the y side
1/1,5 = 4 / x equals that x = 6
I didn't understand why at 1:11 the base was 1.5 units. It seems a little arbitrary to me. Did we just 'infer' it from the diagram?
Radius of the bigger circle, which has a diameter of 3, so half of that. And the radius in this case is the distance between the tangent point "on the bottom" of the circle and the light blue tangent line....
@CalebBlackhand the light blue line is a tangent? I thought it was just a marker line.
@@Bladarc From the way Andy used it throughout the whole solution, it must be a tangent to both circles as well as the height of the triangle....
Is it given that the light blue line is tangent to both circles? I thought it was just the height
My 30 second mental math answer for the next one is 7.5 units squared. We'll see
The next and the last aggvent calendar problem:
If the rectangle width = 2 + a, the diagonal = 6√(a^2 + 1). So we can write
√[(2 + a)^2 + 6^2] = 6√(a^2 + 1)
(2 + a)^2 + 6^2 = 36(a^2 + 1)
35a^2 - 4a - 4 = 0
(7a + 2)(5a - 2) = 0
a = 2/5
Rectangle area = 6(2 + a) = 12 + 12/5 = 14.4
I didn’t read all your solution, but the circles are assumed to be tangent to the vertical lines, Not to each other.
@@JLvatron Thank you very much! I modified my answer. It is now much less messy 😅
If x+2 is the width of the rectangle and 6 is its length, then the diagonal of the rectangle is the hypotenuse of the right triangle that represents half of the rectangle, and the circle drawn inside this triangle has a radius of 1, so the diagonal is equal to x+2+6-2r=x+2+6-2=x+6, so the right triangle has sides x+2,6,x+6, then we write the Pythagorean theorem to determine x.
@@AzouzNacir Whatever you do, find the x 😊
The equation is (x+2)²+6²=(x+6)² and its solution is x=1/2 and the area of the rectangle is (1/2+2)*6=15 and this is the second good catch😅
I'm stumped on tomorrow's! EDIT: Never mind! I found a fairly simple solution using Pythagorean theorem and the rule of intersecting tangents: Let x be the height of the rectangle: 6^2 + x^2 = (5+(x-1))^2... x = 5/2. So rectangle is exactly 15 units squared. :)
It's similar to current. Base is x + 1, height is y + 1, hypotenuse is x + y. Because of circles, base is 6, which gives x == 5, then Pythagorean theorem kicks in and gives y == 1.5, and then 6 x 2.5 == 15.
lemme ask you what is log(-1) and log(i)
How exciting 😃
Next and last problem: Area: 15
How exciting
Nooooooo, its January now 😢😢
It's not clear from the sketch or your introduction that the two circles are tangent to the height.
The base is 6 hardestthing to guess 😂😂😂😂😂😂😂😂😎😎
Only one more to go!!
For last question 12√3?
Looks clever, so I support this answer wholeheartedly.
@KingGamereon Might look clever, but it isn't correct.
@@Grizzly01-vr4pn ok I'll believe anyone
Use intersecting tangents to define a variable that is the same distance from tangent to either yellow or blue circle to one corner of one triangle, then use Pythagoras to solve for that distance. This will give you the height of the triangle thus height of the rectangle thus the area. And it's not 12sqrt3
the base of the big rectangle is 6*(the radius) = 6 ,
using the intersecting teangents : the diagonal is X+Y but X=6-1 =5 so the diagonal is 5+Y ,
the height is Y+1
using the pythagore theorem : 6² + (1 + Y)² = (5 + Y)²
Y² + 2Y + 37 = Y² + 10x + 25
Y = 1.5
so the height is 1+1,5= 2,5
the area is 2,5 * 6= 15
(sorry if english is bad , not my language)
Day 31: 15 sq units
I used a lot of trigonometry and solved it
day31
1.(x+5)^2-(x+1)^2=6^2
→4(2x+6)=36→x=3/2
2.A=6(x+1)=15😊
Well well well
hi
I'm early for once?
As an Indian who dreams of coming to the United States to contribute to its diversity, I deeply admire the values of freedom and opportunity that this nation represents. However, I’d like to understand how native Americans view immigrants and the challenges of racism. Do you think America is welcoming to immigrants, or does racism still pose a significant barrier for those seeking to build a life here?"
first