Well it will apply in shapes with a 90 degree triangle ofc. Squares and rectangles sure, but not trapezoids or kites and such (at least not all of them
I can't remember ever learning that an inscribed triangle along the diameter is a right triangle, but it makes sense. That's the conceptual step I was missing.
@@Zieki99 Not that I can recall, but highschool was more than a decade ago and I don't use geometry in my day-to-day job. Again, it's just something I can't remember ever being covered, not that we didn't cover it.
The last time I needed to solve problems like this was around 15 years ago, but i still come here and try to solve these once in a while. The way you teach and explain is so good! Kudos for not only keeping students challenged, but people like me as well!
Good assumption, I just used the blue square is a little over 2.5 so nearest while number fitting scale being 3 and working through 😅yours is a much better assumption and much cooler too
I’d also assume the other unstated (but visually implied) given was that the arc intersected exactly at the lower left corner of the pink square. There’s nothing that says it is drawn to scale, but I don’t see a way to solve it without that implied corner contact variable. Good stuff.
Tried solving it but without that it's impossible. I saw that it was probably what I was missing but given it wasn't stated in the question you can't assume it's a fact.
This is also how I would approach this problem to begin with, how do you scale purple and blue square so that the third, pink square, will touch both the semicircle and align with top of the blue one? Purple and blue depend on each other (otherwise they either aren't squares, or don't add up to 5 base), so there is only one degree of freedom, then for each pair the pink square is implied and either accepts or rejects the solution.
Ah I see. The entire time I was wondering how it even makes sense to find a unique solution, given that you can draw the other 2 squares for any blue square, but with that restriction that doesn't really hold.
@@chaoticsquidno it's Def solveable without the circle. We know it's Fibonacci sequence. Which means we can run that sequence with a base dummy variable adding up each time and dividing that by five. The dummy variable represents the edge of orange square, and fibonnaci sequence dictates it'll repeat five times by now.
Thank you so much for the great content! As one who works as a math teacher, your content has been a huge inspiration on how to make challenging and fun puzzles!
I never comment. Never subscribe. But you are crushing it. I save all of the problems that involve basic algebra geometry and algebra 2 concepts for my high school students. Andy Math out here differentiating instruction for me. God bless you and your family
Love your videos, doing these things with you are one of my favorite activities. Please take care of your own health and don't overdue with the videos and or any other job. Love you.
I love this channel. Would love to see a good explainer/refresher on exactly how integrals are solved, particularly when doing u substitution with going from dx to du. 😊
Seeing this video reminded of my college days and learning of a square with negative dimensions. I think veritisium made a video about real life shapes that exist as the number i. Great content! Keep it up brother.
You got as much views as your subscriber count in just 20h, wow ! This month has made your channel so viral: you've got your 2 most viewed video just this months. How insane ! Congratulations ! 🎉 Keep up the good work like that! 👍
I have an exam today, and they ask a lot of area type geometry questions, I have been following you for a long time, if I get atleast one question that have concepts that you used, all Credit goes to you❤
Fibonacci and Generalization - Building on your process, paying close attention to the triangle which is the sum of two smaller ones, with specific ratio: We have x. Next, we have y=x+x=2x. Next, we have x+y=2x+x=3x. Next, we have x+2y=3x+2x=5x. This is the Fibonacci sequence where each term is multiplied by x. The first, same value as second, term is missing (we have 1, 2, 3, and 5, instead of 1, 1, 2, 3, and 5 - which correspond to the "bite" missing from the "complete" rectangle). using s for side lengths and a for areas, each followed by 1-3 for smallest to largest squares. Let's call the 4th term: z. specifically: z=5x. so: s1=z/5, s2=2z/5, s3=3z/5. Squaring for area: a1=z²/25, a2=4z²/25, a3=9z²/25. Summing: Total area = 14z²/25. In your example, z=5, so the total area = 14*5²/25=14. For z=6, for example, total area would be 14*6²/25=20.16. For z=10, twice the 5, the result should be quadrupled: 14*10²/25=56 - and it is.
For all those saying, "its obvious" or "i could just visualise it" etc.... That means nothing in an exam as its not a proof. You will get next to 0 points just giving an answer or an explanation like that. You need to PROVE the answer. Yes you can guess but it doesnt mean its right. What if y isnt 2x? Thats why we have proofs.
what if y isn't 2x? (x + 2y) / 5 = 1 assuming 5x = 5 5x - 2y =x y=2x and x=1 let's fill this in x + 2y =5 1 + 2(2) = 5 hrmmmmmm sir I have a suspicion it y may be 2x
Yeah, it can bite you back Especially shapes where they LOOK like they're touching each other so you can visualize and make comparison....but turns out they werent when try to proof it and end up not getting an answer on all 5 options
I understand the process as soon as i see the problem. Math is very interesting, exciting and challenge. I love to go back in time and wanna challenge these math problems again😢
Wow, amazing how these can be solved, when at a first glance it seems impossible! Love watching these to brush up on my math skills👍 Trigonometry is my favourite!👍
Learned something new with your way of solving it. I managed to solve it but I assumed the smaller triangle intersect was right at the middle - Which I think it’s not a given so I’m considering my process luck 😅
With math problems like this a lot of the time they're not to scale, so it's important to check, but it's satisfying that the solution to this one actually is what it looks like.
An alternative way to solve this problem:trace a line beetwen the center of the semicircle and the point where the semicircle intersect the lower left corner of the smaalest square and apply the pitagorean theorem în the right triangle.
I did it a different way, label lengths of the squares from largest to smallest as a,b,c. Then we can create a set of equations a+b = 5 (1) b+c = a (2) To get a third equation we can take the point where the corner a is on the semicircle and use pythagoras, noting that the radius is r = 2.5: b^2 + (2.5 - c-a)^2 = 2.5^2 (3) add together (1) and (2) to get a+c = 5 + a - 2b = 5 + (a+b) - 3b = 10 - 3b (4) substitute (4) into (3) to get b^2 +(3b-7.5)^2 = 2.5^2, ==> 10b^2 - 45b + 50 = 0, ==> (2b - 5)(b - 2) = 0 ==> b = 2, 2.5 if b is 2.5 then a = 2.5 and c = 0 so total area is 12.5 (trivial solution) , and for b = 2, we get a = 3 and c = 1 so total area is 14.
I would like to know which software u use to make these videos (the software in which u teach) cz I wanna start teaching maths after my exams to my juniors through internet
i did this using the Pythagoras formula. you know the center of the circle is 2.5 from the edge. you can draw a triangle that goes from the center of the circle to intersection of the circle and the 2 smaller boxes, then down perpendicular to the base of the semi circle. this has sides 2.5 (hypotenuse is a radius of the circle), y and 2.5-y+x. we can substitute x = 5-2y into that last side then the side lengths into Pythagoras formula to give us a quadratic in y. solve that to get y = 2 or y = 2.5 giving x = 1 or x = 0 (which we can discard) and finish up getting the areas.
I solved using the equation of the circle for 3 points like one of the other video you showed, you can declare 2 variables hx and Xx and you can write a system with 4 incognita and 4 equations solving for hx and Xx you get the same result but in a less elegant fashion
Very nice! However there exists another solution just using the assumptions you made. The equation at 2:13 assumes y is not 0, but for y=0 you do acually get another solution for the total area, being 12.5 square units.
The only way I see that working is if the side lengths of the squares have to be integers. Otherwise, there's ways I can manipulate the squares where pink is ALMOST the size of purple. That gives us an area of approximately 2x by 3x where 3x = 5. Final equation for the area gives us 5 × 5 × 2/3 = 50/3 square units, which is more than 14. Other way around, make pink approach zero. Area is now 2y^2 where y equals 5/2, the answer is 25/2 square units which is less than 14. Any other answers will lie between those extreme values.
@tomdekler9280 I suppose I did also assume that. Good point. I came to the point of, 5=2x+y where x is purple and y is pink. From there I said 2x must be an even number less than 5. this is where I assumed an integer value, it's also where I said pink must be smaller because otherwise purple could be 1.
I defined x as the length of a side of the smallest square like you did. Then I found ways to express the sides of the other squares with using x as the only variable: pink edge = x purple edge = 2.5-0.5x blue edge = 2.5+0.5x Then I put a right triangle between the bottom left corner of the smallest sqaure, the center point of the half circle and somewhere on the base line directly underneath that bottom left corner of the smallest square. The hypothenuse of that triangle would be identical to the radius of the half circle which is 2.5 and the other sides would be 1.5x and 2.5-0.5x (the purple edge). Using the pythagorean theorem I found x (the pink edge) to be 1 and substituting that into the above I found the purple and blue edges to be 2 and 3 respectively. 1²+2²+3²=1+4+9=14
You say x can’t equal 0, but I don’t see why not. You get the pink square with an area of 0, and the purple and blue squares each with an area of 7.75, for a total area of 12.5, making it a rectangle.
I support your comment. x = 0 is a totally legit solution. It shouldn' t be a surprise that a problem that implies a quadratic equation has two solutions.
I solved it a different way. The radius of the semicircle is half the diameter, so r=5/2. Drawing a line from the center of the semicircle's base to the point at which the two smaller squares and the circle meet results in a line of length 5/2. Draw a line straight down from that point. You now have a right triangle with hypotenuse of length 5/2 and a height of y. Thanks to the Pythagorean Theorum, the base of that triangle, which goes from where that vertical line of length y touches the base to the midpoint of the semicircle's base, is the square root of the difference between (5/2)^2 and y^2. So the triangle's sides are sqrt((25/4)-y^2), y and 5/2. The length of the line extending from that triangle's right angle to the left edge of the semicircle is y-x. Therefore, the sum of y-x and sqrt((25/4)-y^2) is equal to the radius, which is 5/2. That's (5/2)=y-x+sqrt((25/4)-y^2) Solve for x, and you get x=y-(5/2)+sqrt((25/4)-y^2). Replace x with that expression in 5=2y+x, and you get 5=3y-(5/2)+sqrt((25/4)-y^2). Isolate the radical to get (15/2)-3y=sqrt((25/4)-y^2). Square both sides to get (225/4)-45y+9y^2=(25/4)-y^2. Move everything over to one side and combine like terms to get the quadratic function 10y^2-45y+50=0. Simplify to 2y^2-9y+10=0. Utilize our old friend, the quadratic formula, to get y=(9+-sqrt((-9)^2-4(2)(10)))/2(2). Simplify: y=(9+-sqrt(81-80))/4. Simplify: y=(9+-sqrt(1))/4. Simplify: y=(9+-1)/4. Conduct plus/minus operation: y=10/4, 8/4. Simplify: y=5/2, 2. y must be less than the radius, which is 5/2. Therefore, y can only be 2. Plug it in to 5=2y+x, solve for x, and x=1. Plug x and y into x^2+y^2+(x+y)^2 to get 14.
Nice problem and nice resolution, bro! I've worked out the second equation by another triangle: the x²+y²=5² and used the the first equation squared as y²=(5-2x)².
Completely forgot about that right angle theorem; there are just so many from Geometry to remember .. I kept thinking about trying to use Pythagorean theorem on the x and y blocks to find the radius of that semi-circle..
Approached this slightly differently. Made a right triangle from the center of the circle to where it intersects the pink and purple squares. One side is y, hypotenuse is 2.5 and the other side is the radius of the circle (2.5) minus the portion of purple square that sticks out past the pink square (y - x). So right triangle with sides y and 2.5 - (y - x) and hypotenuse 2.5. Plugged in 5-2y for x and solved the resulting quadratic form Pythagoras.Not quite as elegant but still worked.
Cool! Here's how I solved it instead: - Grabbed a screenshot - Cropped the figure - Resized it so that the long side equals 500 pixels - Measured all sides, assuming that 1 unit = 100 pixels - Obtained measurements of 2.97, 2.03, and 0.95 units - Rounded up the measurements to the nearest integers, assuming that the author of the problem used integers - Obtained measurements of 3, 2 and 1 - Calculated the area of each square - Summed the areas
For me it's purely visual, I saw that the length was 5 then found that the smallest square is one by one long with an area of 1 and simply counted how many small squares would be present in the shape we are given. Granted this doesn't work in every scenario but I got to the answer a lot quicker than the video does and got a good chuckle when me and the video arrived to the same conclusion.
