Aggvent Calendar Day 10

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  • Опубликовано: 24 дек 2024

Комментарии • 105

  • @username5227
    @username5227 14 дней назад +241

    1:05 never let em know your next move

    • @kmjohnny
      @kmjohnny 14 дней назад +5

      things are getting spicier

    • @ChrisMMaster0
      @ChrisMMaster0 13 дней назад

      Really cought me off guard

  • @true_folu
    @true_folu 14 дней назад +255

    not the ellipse...
    there is our beloved box?...

    • @trumpetbob15
      @trumpetbob15 14 дней назад +16

      Just not the same without a box around it.

    • @frankstrawnation
      @frankstrawnation 14 дней назад +24

      Your question looks important. Let's put a box around it.

    • @eikebehrmann3493
      @eikebehrmann3493 14 дней назад +20

      Ellipses are for interesting things. Boxes are for important things. #BringBackTheBox

  • @Kyuko-chan246
    @Kyuko-chan246 14 дней назад +115

    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

  • @MichaelMeinberg
    @MichaelMeinberg 14 дней назад +66

    Everytime you solve one of these types of problems without solving for the individual variables, it just blows me away.

    • @konradyearwood5845
      @konradyearwood5845 11 дней назад

      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.

  • @rmichaeldeutsch
    @rmichaeldeutsch 14 дней назад +132

    Did we even solve it if we didn't put a box around it? 🤔 #bringbackthebox #howexciting

  • @esrAsnataS
    @esrAsnataS 14 дней назад +21

    Never ceases to amaze me how you can get answers from such little information.

  • @vortexlegend101
    @vortexlegend101 14 дней назад +36

    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

    • @charimonfanboy
      @charimonfanboy 14 дней назад +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

    • @DieterKramer-ju3mj
      @DieterKramer-ju3mj 14 дней назад +1

      Y is in fact zero

    • @egg_948
      @egg_948 13 дней назад +2

      ​@@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

    • @atrus3823
      @atrus3823 13 дней назад +4

      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.

    • @wolfgangrenz3693
      @wolfgangrenz3693 13 дней назад

      Since both extremes are identical, proof requires still that it is independent of the angle or independent of x.😮

  • @fieryweasel
    @fieryweasel 13 дней назад +17

    "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..."

  • @5gearz
    @5gearz 13 дней назад +6

    I felt a shiver down my spine when I didn’t hear box

  • @pedroamaral7407
    @pedroamaral7407 14 дней назад +12

    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
      @crosswingrobots 14 дней назад +3

      I used your second method. Whenever I see an underconstrained geometry problem I just solve it at one of its extremes.

    • @pedroamaral7407
      @pedroamaral7407 14 дней назад +1

      @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)

  • @rowrow_
    @rowrow_ 13 дней назад +3

    The elegant solutions you make really remind me how much I enjoy math. Thanks Andy

  • @stevekilligrew788
    @stevekilligrew788 8 дней назад

    How cool is school! Many Thanks & best wishes to you for a very Merry Chrimbo & a Happy New Bodmas 2025!🇬🇧

  • @pavelpotocek
    @pavelpotocek 6 дней назад

    Much simpler answer: Imagine the picture showing the marked line (close to) horizontal. That's clearly also a valid diagram. Then, the circle diameter is 6. The rectangle area is then diameter times radius, which is 6*3=18.

  • @divVerent
    @divVerent 10 дней назад

    Biggest hint is the lack of any orher numbers in the problem. Makes you immediately think of there being some "hidden facts" to use.
    Let's call the two points on the line of length 6 B and C, and the other endpoint of the half circle A. Then ABC is a right angled triangle (Thales). The left bottom point of the rectangle shall be called H, so that HC is a height of the triangle.
    If you remember your Pythagoras proof, you remember that HB * AB = BC^2 (aka q * c = a^2), and you win. If not, observe that ABC is similar to CBH due to having the same angles (which follows from the triangle angle sum being 180° and the known angles alone). From that you get BC : AB = HB : BC and after multiplying out, BC^2 = HB * AB.
    And as our rectangle is just HB * (AB / 2), we know the rectangle area is 18.

  • @TurquoizeGoldscraper
    @TurquoizeGoldscraper 13 дней назад +1

    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.

    • @geoffreycleasby1939
      @geoffreycleasby1939 11 дней назад

      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!

  • @paparmar
    @paparmar 13 дней назад +1

    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.

  • @JohnDoe-we
    @JohnDoe-we 14 дней назад +5

    I love you andy math

  • @jimlocke9320
    @jimlocke9320 13 дней назад

    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.

  • @andrejradovic2124
    @andrejradovic2124 13 дней назад

    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 (*) )!

  • @atrus3823
    @atrus3823 13 дней назад

    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.

  • @Kero-zc5tc
    @Kero-zc5tc 4 дня назад

    You can solve it much faster, like 20 seconds fast. The angle between the base and the line ‘6’ can be seen using logic as irrelevant (the angle does not affect the answer), so set this angle equal to 0 and find the base of the rectangle is 6 and the height is the semicircle radius which is 3, so 6x3=18.

  • @SumbudysDad
    @SumbudysDad 11 дней назад +1

    And I thought I was a nerd for watching the entire video! Guess ima super nerd for coming to read the comments too!!!!😂😂

  • @benjaminmichael5719
    @benjaminmichael5719 14 дней назад +5

    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.

    • @Qermaq
      @Qermaq 14 дней назад

      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.

    • @RylanceStreet
      @RylanceStreet 14 дней назад

      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.

    • @Qermaq
      @Qermaq 14 дней назад

      @@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.

    • @Epyxoid
      @Epyxoid 13 дней назад

      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!

    • @myusualnickwastaken
      @myusualnickwastaken 13 дней назад

      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.

  • @loduwk3
    @loduwk3 11 дней назад +1

    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

  • @alanderson78
    @alanderson78 13 дней назад

    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.

  • @feyh
    @feyh 14 дней назад +3

    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.

    • @voliol8070
      @voliol8070 13 дней назад

      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.

    • @jimi02468
      @jimi02468 13 дней назад +1

      @@voliol8070 The notorious "proof by assuming a unique answer" which, although is right, is cheating.

  • @BowieZ
    @BowieZ 13 дней назад

    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.

  • @michaelbellefroid7736
    @michaelbellefroid7736 14 дней назад

    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?

  • @Insightfill
    @Insightfill 13 дней назад

    Golden! Never needed to actually solve for r or x!

  • @Sina-dv1eg
    @Sina-dv1eg 14 дней назад +1

    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.

  • @CharlesB147
    @CharlesB147 14 дней назад

    That was a really clean solution.

  • @electricsleep
    @electricsleep 14 дней назад +1

    My guess for day 11 is 135° not certain on it though

  • @tedforringer9124
    @tedforringer9124 11 дней назад

    So good!

  • @nessleepk
    @nessleepk 13 дней назад

    I liked the answer floating up and erasing the work, if only it was kinda square shaped like an eraser

  • @thetalkativecrow
    @thetalkativecrow 13 дней назад

    Love this one.

  • @coogs5688
    @coogs5688 12 дней назад

    Ellipse? Kid's crying now, thanks

  • @TazPessle
    @TazPessle 14 дней назад +1

    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.

  • @gdash.dasher-y7e
    @gdash.dasher-y7e 13 дней назад

    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

  • @charimonfanboy
    @charimonfanboy 13 дней назад

    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

  • @MichaelDarrow-tr1mn
    @MichaelDarrow-tr1mn 11 дней назад

    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

  • @heathadams6913
    @heathadams6913 12 дней назад

    Woah, why put an "eclipse around it"? How inciting!

  • @Doc_Fartens
    @Doc_Fartens 13 дней назад

    I *almost* stopped watching at that cursed elipse. How could you do this to us?! Where is our beloved box?!

  • @joeschmo622
    @joeschmo622 13 дней назад

    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. 😂

  • @JasonMoir
    @JasonMoir 14 дней назад

    An ellipse? You mad lad....next you'll be using a rhombus.

  • @EinChris75
    @EinChris75 14 дней назад

    So box around a solution? Why did he not solve that puzzle. Puzzling. And exciting.

  • @al3h4ndr05
    @al3h4ndr05 14 дней назад +1

    We are on day 11 already

    • @Memetist
      @Memetist 14 дней назад

      Yeah right 😂

  • @pekkokerokoski
    @pekkokerokoski 14 дней назад

    These are so fun

  • @anathmaity2000
    @anathmaity2000 4 дня назад

    Very nice sir

  • @erenerbay
    @erenerbay 14 дней назад

    NOOO WHY DID YOU PUT AN ELLIPSE 😭😭😭😭

  • @Doggo2_cool
    @Doggo2_cool 13 дней назад

    Day 10 of asking where you solve your problems. Like in google docs?

  • @MathTrick-t7v
    @MathTrick-t7v 4 дня назад

    Amazing

  • @leahithink
    @leahithink 13 дней назад

    no... no box?

  • @KidFifty
    @KidFifty 14 дней назад

    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.

  • @awtrfghadfet2200
    @awtrfghadfet2200 10 дней назад

    I think it may not be the correct answer. If it is 18 it would mean height 3 times base 6. And the base is a bit shorter than 6...

  • @arbuzow
    @arbuzow 14 дней назад

    I'm in shambles... How could you use an ellipse? I'm literally shaking rn. Not exciting at all.

  • @jimi02468
    @jimi02468 13 дней назад

    It took me literally five seconds to solve this.

  • @gutwrenchingdeaths
    @gutwrenchingdeaths 13 дней назад

    I haven't watched the video yet, I solved it but I feel like I over complicated it

  • @newdenzellplayz
    @newdenzellplayz 13 дней назад

    WHERES THE BOX

  • @tellerhwang364
    @tellerhwang364 13 дней назад

    day11
    180-90/2=135😊

  • @nathanc6516
    @nathanc6516 14 дней назад

    Not sure if I'm right, but I think tomorrow's answer is 135 degrees.

  • @thoperSought
    @thoperSought 13 дней назад

    an elipse?!???!?!!!?!!?11?!?

  • @Piggels
    @Piggels 13 дней назад

    wheres my box..

  • @peteevans7776
    @peteevans7776 14 дней назад

    bac colors, yellow on white is almost impossible.

  • @stevechrisman3185
    @stevechrisman3185 13 дней назад

    Ellipse ? ELLIPSE ??? What's Happening ???????????????????????????

  • @weeklyhan8506
    @weeklyhan8506 14 дней назад

    1:05 he is slowly losing it

  • @5gearz
    @5gearz 13 дней назад

    Your a day behind! Noooo!

  • @laserrrrr
    @laserrrrr 14 дней назад +2

    hi andy can you make a video about derivatives and functions? im in grade 8 and i find math rly interesting

    • @SLift-mn8hi
      @SLift-mn8hi 14 дней назад

      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.

    • @blintzy6969
      @blintzy6969 14 дней назад

      Same but it’s called Year 9 in UK

    • @laserrrrr
      @laserrrrr 13 дней назад +1

      @@SLift-mn8hi oh ok imma check that out thanks

  • @Xuelynom
    @Xuelynom 13 дней назад

    No you
    Can't
    Solve it
    Assuming
    Constant
    Solution !

  • @wasimahmad-t6c
    @wasimahmad-t6c 6 дней назад

    5.7÷2=2.85×5.7=16.2

  • @ethanfreel1222
    @ethanfreel1222 14 дней назад +9

    Dislike. No box around the answer.

  • @rayyanrehan98
    @rayyanrehan98 14 дней назад +1

    First

  • @fjaps
    @fjaps 14 дней назад

    if you don't bring the box back next video i might unsubscribe

  • @israelgonzalezsosa1520
    @israelgonzalezsosa1520 14 дней назад

    Hardest problem ever dislike

  • @zobiaibrahim
    @zobiaibrahim 14 дней назад

    Hi Andy, I have a problem that I want you to solve. I can send it on your email? And what is it?

    • @richardl6751
      @richardl6751 14 дней назад

      You can contact him through his web site.

  • @ekoi1995
    @ekoi1995 11 дней назад

    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.