That is the beauty of Algebra. When I was doing maths regularly at this level finding the answer was often anti-climatic as a lot of the fun was derived from within the challenge of determining and deciding how to manipulate the variables to derive the solution. I really enjoy this channel.
A quicker solution if you allow some assumptions: Since we are not told what angle the line of length 6 units is at, we can assume that if the question can be answered, the answer won’t depend on the angle. Therefore if the angle can be anything, make it zero degrees, so the rectangle just encloses the semicircle, this means the diameter is 6 and thus the radius is 3, so the rectangle’s area is 6x3 = 18 units^2
I did pretty much the same, but couldn't prove it was true for any angle so only gave myself half a mark. Also, I can't seem to do the same for the tomorrows one since if the blue circle was the same size as the yellow circle, the angle would be 135 degrees while if the blue circle was small enough for the rectangle to be a square, the angle would be 152.5 degrees. And I can't see anything do define any sizes or ratios
Wow, this is slick. Another way you could do this is the other extreme: 45°. Then area is just r^2. 6^2 = 2r^2 from Pythagoras so r^2 = 18. Not quite a quick, but still satisfying.
"Hi, I'm calling from Consolidated Boxes about your recent order. I can only assume there was a mistake in the paperwork, since I see you ordered an ellipse..."
I did it in an easier way, I think. I connected the left end of the semicircle to the secant point. This will form a right triangle (leg 6 and hypotenuse 2r) similar to the rectangle with base x+r, height y. So I made the similarity: 6/2r = b/6 (where b is the base of the rectangle). Therefore, b*r = 18 (which is exactly the area of the rectangle). It is interesting to note that for any size of rectangle and semicircle, the area remains the same. So you could suppose a semicircle exactly the size of the base of the rectangle and you would get the same answer.
@crosswingrobots In a test I would do it just like that too, but I wanted to solve it in a more traditional way for fun lol (and to see if I could get the same result)
If we make the assumption that it has a constant solution and that the angle between the bottom of the rectangle and the length of 6 doesn't matter, then we can simplify this by making the 6 line up with the bottom. This gives us a semi-circle with diameter 6, which is a radius of 3, which is also the height of the rectangle. Therefore the rectangle has width 6 and height 3, and an area of 18.
At the same time, you can make the rectangle base shorter, to just reach the centre of the circle; the rectangle becomes a square, radius by radius, with diagonal length 6; lo and behold, the area is still 18!
So the area of the rectangle will always be D^2 / 2, where D is the length of the "diagonal" (6 in this video). Can check this using limiting cases: if x = 0, then the rectangle is a square of side r, and D will be r*SQRT(2), and D^2 / 2 will be r^2, the area of the square. If x = r, the diagonal becomes the same as the base of the rectangle (i.e., 2r), whose area is 2 * r^2, which is also (2r)^2 / 2.
If the semicircle is divided into left and right quarter circles, the problem statement implies that one point on the left side of the rectangle lies on the left quarter circle and the rectangle area is the same no matter where the point lies on that quarter circle. Andy Math solves the general case, but there are two special cases that are straightforward to solve. First. let the intersection point lie on the diameter of the semicircle. Then, the diameter is 6 and is the base of the rectangle, so the radius is 3 and the height of the rectangle, the area is bh = (6)(3) = 18. Second, let the intersection lie at the peak of the semicircle. The line of length 6 and the diameter of the semicircle form an inscribed angle intercepting a 90° arc, so the inscribed angle is 45°. The left side of the rectangle is a radius of the semicircle. An isosceles right triangle is formed with hypotenuse 6. The sides are 3√2. The area is (3√2)² = (9)(2) = 18. On a multiple choice test, mark 18 units² as correct and go on to the next problem. If you aren't required to show your work, give the answer as 18 units². If you are required to show your work, proceed to solve the general case. The answer 18 units² for the two special cases will serve as a check that you have solved the problem correctly.
An alternative solution for those familiar with right triangles: Let us draw a right triangle with sides a and b enclosing the right angle and hypotenuse c in such a way that c is horizontal in the bottom of the figure and the right angle is on top of the figure. Dropping the height h from the right angle to the hypotenuse c, which is then subdivided by the height into two parts, namely m adjacent to side a and n adjacent to side b so that c = m + n, we then have a^2 = m c and b^2 = n c . (*) (Also: h^2 = m n.) (This is easily proved by means of similar triangles, for example, the triangle formed by a, h and m is similar to the overall triangle with corresponding sides c, b and a, giving a : m = c : a; similarly we obtain b : n = c : b, and also h : m = n : h. And by the way: the Pythagoras theorem a^2 + b^2 = c^2 results immediately from the formulas (*) by adding the left-hand and right-hand sides! It is useful to be able to recall these formulas without hesitation.) In the problem above our right triangle is within the Thales (semi-)circle with a = 6 and c = 2r. Using the first formula in (*) gives a^2 = m 2r, and as the sought-after area A equals A = mr, we obtain immediately A = a^2 / 2 = 18. Quick and simple (knowing (*) )!
This is one of those that, no matter if the base of the rectangle is equal to the radius or the diameter of the semicircle or anything in the between, the answer would be 18. I solved it just tryin the two edge cases.
I did it a simpler way, which I think might count as cheating lol. Given that the length 6 is our only given number, ie. we don't have the circle's radius or rectangle's length, we can infer one thing: As long as the line connecting the two intersections between the circle and the rectangle's vertical sides is equal to 6, then the solution will be the same even if we change the circle's radius or the rectangle's length. For example, if we increase the circle's radius (and thus, the rectangle's height), the line would become longer, so we'll need to make the rectangle narrower to keep it at 6. So, now we can just use the simplest example. Let's widen the rectangle so it goes all the way to the end of the circle. That way, the line with length 6 becomes equal to the rectangle's length and circle's diameter. And since the rectangle's height is equal to the circle's radius, the area of the rectangle is 3*6=18.
Tomorrow's one has me stumped. I can work out the answer by adjusting the circles to be the same size... but have no idea where to start proving it for other sized circles.
I thought of the figure flipped across y axis. Lower-left corner of rect at origin. Then w = 6 cos theta and h = r. What's r? To find that I considered the circle centered at (r, 0) to align with rectangle: r^2 = (x - r)^2 + y^2. Plug in known point (6 cos theta, 6 sin theta). A lot of stuff cancels and you get r = 6/(2 cos theta). Multiply by w = 6 cos theta, cos thetas cancel and you have 18. Not the simplest, but was the first thing I thought of.
Tomorrow's puzzle looks insane. Edit: I was lost on how to tackle this one. So for the first time in the aggvent series, I sketched up the problem in CAD. I added all the known (and assumed) constraints to see if I could fully constrain the puzzle so that there's a fixed angle. I was unable to fully constrain every element in the puzzle. So I was able to drag the diameter of the smaller circle, and it would give a variable angle. The angle between the two line segments is dependent on the size of the blue circle in relation to the yellow circle. The yellow circle is fully constrained because it's tangent to 3 sides of the rectangle. We can assume that the line segment in the yellow circle is coincident with the point where the yellow circle meets the top part of the rectangle. The other side of the line segment in the yellow circle meets at the tangent point where the yellow and blue circle meet, and a tangent line between the two is not vertical (otherwise both circles would be the same size). The blue circle's diameter is also not horizontal, because the point where the angle is being asked, must be coincident. Also the blue circle would have to be the same size as the yellow one, for the line segment to be perpendicular to the blue circle's tangent to the rectangle. So based on how I set up the CAD sketch, it needs an additional constraint to solve for the angle. I'm looking forward to see what mathematical magic you pull off to solve this one.
Remember that if an angle from a point on a circle to a chord (the point is not inside the chord part) subtends an angle measuring A, the angle to that chord from the center is 2A.
Yes you need an additional constraint to solve for your angle, but you don't to solve for the area of the rectangle. The solution given doesn't produce specific values of x, y and r - these can vary, but the area remains constant.
@@RylanceStreet This is a fairly simple trig problem, it relies on a well-known identity, but I suspect he will do a geometric solution. If we connect the centers and draw radii to the left of the small circle's center r and at the top of the large circle R, we get two isosceles triangles that share a relationship with the right triangle with legs R - r, an unknown distance, and hypotenuse R + r.
I'm not entirely sure about how you reconstructed this problem. Yes, the yellow circle is fully constrained, so as the blue one, since it's tangent to 2 sides of the rectangle and to the yellow circle. Also, the line segment in the blue circle is not a diameter!
There is a degree of freedom in the diagram (the radius of the blue circle and the width of the box constrain each other, but they aren't constrained by the rest) but the angle is the same in every case. That's true in many Catriona Agg problems (including this one #10) and actually gives another way to solve them: set the unconstrained parameter such that the problem becomes easier; since all these problems have a single constant as an answer, you can solve any constrained version of them and get the same answer as in the general case.
One thing I have learned from these videos: I it looks like there is not enough information to solve, there is. Move the left side of the rectangle to the right to make a square with a diagonal of 6.
Damn, I jumped on my chair hearing that "let's put an ellipse around it"! @AndyMath: I'm curious to know, your solutions are always clear and straight to the point. But how long do you take to actually find a solution to a problem like this one?
Recognising that the angle is a free variable, i asked what happens as the angle approaches zero. A 3 x 6 tectangle, area 18. Same value for angle of 45° with a √18 by √18 square. Edit: the angle at the lower rhs.
135 degrees, possibly call the angle we need a draw a line between the two centres, this is a straight linethe angle between angle a and the blue circle's radius call b and the angle between angle a and the yellow circle's radius call c draw another radius from the centre of the blue circle to the other side of the blue circle's chord to create an isosceles triangle with the outer angles b and the angle around the centre of the circle call e do the same for the yellow circle to form another isosceles triangle with the outer angles c and call the inner angle d draw another line from the centre of the blue circle straight up and a line from the centre of the yellow circle horizontally left, this forms a right angled triangle, call the angle of this triangle in the blue circle f and the angle in the yellow circle g f+g=90 e=90+f d=90+g b=(180-e)/2 c=(180-d)/2 a=180-(b+c) substitute everything in and jigger them about and you end up with a=135 And if you struggled to follow that, I did too
18 square units proof: it can't be not enough information because finding that wouldn't make the video this exact length and if it's anything it's 18 square units, by the case where the length is the diameter of the semicircle
Given that no angle, etc,. was given, I assumed that it wouldn't make any difference, so I just aimed that diagonal perfectly horizontal. So now 2r (horizontal) = 6, r (vertical) must then = 3, so 3*6=18. I couldn't get arsed to formally prove it, though. 😂
I don't know if it's exactly what you're looking for/if you've seen it already, but he posted a video 2 months ago titled "Introduction to Calculus (Derivatives)" that might be helpful.
1:05 never let em know your next move
things are getting spicier
Really cought me off guard
not the ellipse...
there is our beloved box?...
Just not the same without a box around it.
Your question looks important. Let's put a box around it.
Ellipses are for interesting things. Boxes are for important things. #BringBackTheBox
Did we even solve it if we didn't put a box around it? 🤔 #bringbackthebox #howexciting
You scared me so much when you put an ellipse around the answer, I need to watch another video where you put the answer in a box to recover from that
Everytime you solve one of these types of problems without solving for the individual variables, it just blows me away.
That is the beauty of Algebra. When I was doing maths regularly at this level finding the answer was often anti-climatic as a lot of the fun was derived from within the challenge of determining and deciding how to manipulate the variables to derive the solution. I really enjoy this channel.
A quicker solution if you allow some assumptions:
Since we are not told what angle the line of length 6 units is at, we can assume that if the question can be answered, the answer won’t depend on the angle.
Therefore if the angle can be anything, make it zero degrees, so the rectangle just encloses the semicircle, this means the diameter is 6 and thus the radius is 3, so the rectangle’s area is 6x3 = 18 units^2
I did pretty much the same, but couldn't prove it was true for any angle so only gave myself half a mark.
Also, I can't seem to do the same for the tomorrows one since if the blue circle was the same size as the yellow circle, the angle would be 135 degrees while if the blue circle was small enough for the rectangle to be a square, the angle would be 152.5 degrees. And I can't see anything do define any sizes or ratios
Y is in fact zero
@@charimonfanboy one way to think about it is that in the formula for the area (xr+r^2), it doesn't depend on y, which means y can equal 0
Wow, this is slick. Another way you could do this is the other extreme: 45°. Then area is just r^2. 6^2 = 2r^2 from Pythagoras so r^2 = 18. Not quite a quick, but still satisfying.
Since both extremes are identical, proof requires still that it is independent of the angle or independent of x.😮
Never ceases to amaze me how you can get answers from such little information.
"Hi, I'm calling from Consolidated Boxes about your recent order. I can only assume there was a mistake in the paperwork, since I see you ordered an ellipse..."
I did it in an easier way, I think.
I connected the left end of the semicircle to the secant point. This will form a right triangle (leg 6 and hypotenuse 2r) similar to the rectangle with base x+r, height y.
So I made the similarity: 6/2r = b/6 (where b is the base of the rectangle). Therefore, b*r = 18 (which is exactly the area of the rectangle).
It is interesting to note that for any size of rectangle and semicircle, the area remains the same. So you could suppose a semicircle exactly the size of the base of the rectangle and you would get the same answer.
I used your second method. Whenever I see an underconstrained geometry problem I just solve it at one of its extremes.
@crosswingrobots In a test I would do it just like that too, but I wanted to solve it in a more traditional way for fun lol (and to see if I could get the same result)
The elegant solutions you make really remind me how much I enjoy math. Thanks Andy
I felt a shiver down my spine when I didn’t hear box
im not sure if my aproach is right but it doesnt say that it is drawn to scale so the 6 is the circles diameter and 6 times 3 is also 18
If we make the assumption that it has a constant solution and that the angle between the bottom of the rectangle and the length of 6 doesn't matter, then we can simplify this by making the 6 line up with the bottom. This gives us a semi-circle with diameter 6, which is a radius of 3, which is also the height of the rectangle. Therefore the rectangle has width 6 and height 3, and an area of 18.
At the same time, you can make the rectangle base shorter, to just reach the centre of the circle; the rectangle becomes a square, radius by radius, with diagonal length 6; lo and behold, the area is still 18!
So the area of the rectangle will always be D^2 / 2, where D is the length of the "diagonal" (6 in this video). Can check this using limiting cases: if x = 0, then the rectangle is a square of side r, and D will be r*SQRT(2), and D^2 / 2 will be r^2, the area of the square. If x = r, the diagonal becomes the same as the base of the rectangle (i.e., 2r), whose area is 2 * r^2, which is also (2r)^2 / 2.
If the semicircle is divided into left and right quarter circles, the problem statement implies that one point on the left side of the rectangle lies on the left quarter circle and the rectangle area is the same no matter where the point lies on that quarter circle. Andy Math solves the general case, but there are two special cases that are straightforward to solve. First. let the intersection point lie on the diameter of the semicircle. Then, the diameter is 6 and is the base of the rectangle, so the radius is 3 and the height of the rectangle, the area is bh = (6)(3) = 18. Second, let the intersection lie at the peak of the semicircle. The line of length 6 and the diameter of the semicircle form an inscribed angle intercepting a 90° arc, so the inscribed angle is 45°. The left side of the rectangle is a radius of the semicircle. An isosceles right triangle is formed with hypotenuse 6. The sides are 3√2. The area is (3√2)² = (9)(2) = 18.
On a multiple choice test, mark 18 units² as correct and go on to the next problem. If you aren't required to show your work, give the answer as 18 units². If you are required to show your work, proceed to solve the general case. The answer 18 units² for the two special cases will serve as a check that you have solved the problem correctly.
I love you andy math
An alternative solution for those familiar with right triangles: Let us draw a right triangle with sides a and b enclosing the right angle and hypotenuse c in such a way that c is horizontal in the bottom of the figure and the right angle is on top of the figure. Dropping the height h from the right angle to the hypotenuse c, which is then subdivided by the height into two parts, namely m adjacent to side a and n adjacent to side b so that c = m + n, we then have
a^2 = m c and b^2 = n c . (*) (Also: h^2 = m n.)
(This is easily proved by means of similar triangles, for example, the triangle formed by a, h and m is similar to the overall triangle with corresponding sides c, b and a, giving a : m = c : a; similarly we obtain b : n = c : b, and also h : m = n : h. And by the way: the Pythagoras theorem a^2 + b^2 = c^2 results immediately from the formulas (*) by adding the left-hand and right-hand sides! It is useful to be able to recall these formulas without hesitation.)
In the problem above our right triangle is within the Thales (semi-)circle with a = 6 and c = 2r. Using the first formula in (*) gives a^2 = m 2r, and as the sought-after area A equals A = mr, we obtain immediately A = a^2 / 2 = 18. Quick and simple (knowing (*) )!
This is one of those that, no matter if the base of the rectangle is equal to the radius or the diameter of the semicircle or anything in the between, the answer would be 18. I solved it just tryin the two edge cases.
How would you prove that it's constant though? Our answer *could* turn out as a function of x which isn't the same for every x.
@@voliol8070 The notorious "proof by assuming a unique answer" which, although is right, is cheating.
I did it a simpler way, which I think might count as cheating lol.
Given that the length 6 is our only given number, ie. we don't have the circle's radius or rectangle's length, we can infer one thing:
As long as the line connecting the two intersections between the circle and the rectangle's vertical sides is equal to 6, then the solution will be the same even if we change the circle's radius or the rectangle's length. For example, if we increase the circle's radius (and thus, the rectangle's height), the line would become longer, so we'll need to make the rectangle narrower to keep it at 6.
So, now we can just use the simplest example. Let's widen the rectangle so it goes all the way to the end of the circle. That way, the line with length 6 becomes equal to the rectangle's length and circle's diameter. And since the rectangle's height is equal to the circle's radius, the area of the rectangle is 3*6=18.
Tomorrow's one has me stumped. I can work out the answer by adjusting the circles to be the same size... but have no idea where to start proving it for other sized circles.
I thought of the figure flipped across y axis. Lower-left corner of rect at origin. Then w = 6 cos theta and h = r. What's r? To find that I considered the circle centered at (r, 0) to align with rectangle: r^2 = (x - r)^2 + y^2. Plug in known point (6 cos theta, 6 sin theta). A lot of stuff cancels and you get r = 6/(2 cos theta). Multiply by w = 6 cos theta, cos thetas cancel and you have 18. Not the simplest, but was the first thing I thought of.
Golden! Never needed to actually solve for r or x!
My guess for day 11 is 135° not certain on it though
Tomorrow's puzzle looks insane.
Edit: I was lost on how to tackle this one. So for the first time in the aggvent series, I sketched up the problem in CAD. I added all the known (and assumed) constraints to see if I could fully constrain the puzzle so that there's a fixed angle. I was unable to fully constrain every element in the puzzle. So I was able to drag the diameter of the smaller circle, and it would give a variable angle.
The angle between the two line segments is dependent on the size of the blue circle in relation to the yellow circle. The yellow circle is fully constrained because it's tangent to 3 sides of the rectangle. We can assume that the line segment in the yellow circle is coincident with the point where the yellow circle meets the top part of the rectangle. The other side of the line segment in the yellow circle meets at the tangent point where the yellow and blue circle meet, and a tangent line between the two is not vertical (otherwise both circles would be the same size). The blue circle's diameter is also not horizontal, because the point where the angle is being asked, must be coincident. Also the blue circle would have to be the same size as the yellow one, for the line segment to be perpendicular to the blue circle's tangent to the rectangle.
So based on how I set up the CAD sketch, it needs an additional constraint to solve for the angle.
I'm looking forward to see what mathematical magic you pull off to solve this one.
Remember that if an angle from a point on a circle to a chord (the point is not inside the chord part) subtends an angle measuring A, the angle to that chord from the center is 2A.
Yes you need an additional constraint to solve for your angle, but you don't to solve for the area of the rectangle. The solution given doesn't produce specific values of x, y and r - these can vary, but the area remains constant.
@@RylanceStreet This is a fairly simple trig problem, it relies on a well-known identity, but I suspect he will do a geometric solution. If we connect the centers and draw radii to the left of the small circle's center r and at the top of the large circle R, we get two isosceles triangles that share a relationship with the right triangle with legs R - r, an unknown distance, and hypotenuse R + r.
I'm not entirely sure about how you reconstructed this problem. Yes, the yellow circle is fully constrained, so as the blue one, since it's tangent to 2 sides of the rectangle and to the yellow circle. Also, the line segment in the blue circle is not a diameter!
There is a degree of freedom in the diagram (the radius of the blue circle and the width of the box constrain each other, but they aren't constrained by the rest) but the angle is the same in every case. That's true in many Catriona Agg problems (including this one #10) and actually gives another way to solve them: set the unconstrained parameter such that the problem becomes easier; since all these problems have a single constant as an answer, you can solve any constrained version of them and get the same answer as in the general case.
Love this one.
One thing I have learned from these videos: I it looks like there is not enough information to solve, there is.
Move the left side of the rectangle to the right to make a square with a diagonal of 6.
That was a really clean solution.
Ellipse? Kid's crying now, thanks
Woah, why put an "eclipse around it"? How inciting!
Damn, I jumped on my chair hearing that "let's put an ellipse around it"! @AndyMath: I'm curious to know, your solutions are always clear and straight to the point. But how long do you take to actually find a solution to a problem like this one?
Recognising that the angle is a free variable, i asked what happens as the angle approaches zero. A 3 x 6 tectangle, area 18. Same value for angle of 45° with a √18 by √18 square. Edit: the angle at the lower rhs.
So box around a solution? Why did he not solve that puzzle. Puzzling. And exciting.
So good!
I liked the answer floating up and erasing the work, if only it was kinda square shaped like an eraser
135 degrees, possibly
call the angle we need a
draw a line between the two centres, this is a straight linethe angle between angle a and the blue circle's radius call b and the angle between angle a and the yellow circle's radius call c
draw another radius from the centre of the blue circle to the other side of the blue circle's chord to create an isosceles triangle with the outer angles b and the angle around the centre of the circle call e
do the same for the yellow circle to form another isosceles triangle with the outer angles c and call the inner angle d
draw another line from the centre of the blue circle straight up and a line from the centre of the yellow circle horizontally left, this forms a right angled triangle, call the angle of this triangle in the blue circle f and the angle in the yellow circle g
f+g=90
e=90+f
d=90+g
b=(180-e)/2
c=(180-d)/2
a=180-(b+c)
substitute everything in and jigger them about and you end up with a=135
And if you struggled to follow that, I did too
An ellipse? You mad lad....next you'll be using a rhombus.
I *almost* stopped watching at that cursed elipse. How could you do this to us?! Where is our beloved box?!
Guys im a firm eclipse rights believer, i think we should be open to change and not think one thing is < another just because of its looks
18 square units
proof:
it can't be not enough information because finding that wouldn't make the video this exact length
and if it's anything it's 18 square units, by the case where the length is the diameter of the semicircle
We are on day 11 already
Yeah right 😂
These are so fun
Given that no angle, etc,. was given, I assumed that it wouldn't make any difference, so I just aimed that diagonal perfectly horizontal. So now 2r (horizontal) = 6, r (vertical) must then = 3, so 3*6=18.
I couldn't get arsed to formally prove it, though. 😂
NOOO WHY DID YOU PUT AN ELLIPSE 😭😭😭😭
I solved it by using similar right triangles, one which had the diameter as the hypotenuse and one in which the hypotenuse was 6 units.
And I thought I was a nerd for watching the entire video! Guess ima super nerd for coming to read the comments too!!!!😂😂
Day 10 of asking where you solve your problems. Like in google docs?
no... no box?
It took me literally five seconds to solve this.
WHERES THE BOX
I haven't watched the video yet, I solved it but I feel like I over complicated it
day11
180-90/2=135😊
Ellipse ? ELLIPSE ??? What's Happening ???????????????????????????
Not sure if I'm right, but I think tomorrow's answer is 135 degrees.
I'm in shambles... How could you use an ellipse? I'm literally shaking rn. Not exciting at all.
an elipse?!???!?!!!?!!?11?!?
Your a day behind! Noooo!
wheres my box..
bac colors, yellow on white is almost impossible.
No you
Can't
Solve it
Assuming
Constant
Solution !
1:05 he is slowly losing it
Dislike. No box around the answer.
hi andy can you make a video about derivatives and functions? im in grade 8 and i find math rly interesting
I don't know if it's exactly what you're looking for/if you've seen it already, but he posted a video 2 months ago titled "Introduction to Calculus (Derivatives)" that might be helpful.
Same but it’s called Year 9 in UK
@@SLift-mn8hi oh ok imma check that out thanks
First
if you don't bring the box back next video i might unsubscribe
Hardest problem ever dislike
Hi Andy, I have a problem that I want you to solve. I can send it on your email? And what is it?
You can contact him through his web site.
I don't know how to prove it but the Area is ∫₀⁶ x dx = 18. It works on any value for the length of the chord.