Looking back through the "Book of Andy", we had a video (like a year ago) showing the inscribed circle in a 3u-4u-5u triangle has radius 1 (and so area pi). Here, once we arrive at the 3 similar triangles being of side proportions 3-4-5 (@1:47), we can just note the radius of the inscribed circle (say of the largest triangle) to quickly determine the actual dimensions involved. As we have 2.5 units instead of 1, we need to scale a 3u-4u-5u triangle by factor of 2.5 to get the large triangle. Our large triangle must therefore be 7.5u-10u-12.5u, meaning the rectangle is 7.5u x 10u or 75 sq. units.
I solved this slightly differently starting at 2:24, with no further dissection required. You need to draw this to follow, but it's pretty easy. Label the vertices of the triangle A, B, and C starting at the top and going counter-clockwise. Label the points of tangency to the circle A' , B' , and C' with each of these opposite the corresponding vertex of the triangle. Label the center of the circle O. Now, since OC'BA' is a square and AB = 4x, AC' is (4x - 2.5). We know that the distances from a point outside a circle to the tangent points on the circle are equal. So AB' is also (4x - 2.5). Similarly, CB' and CA' are each = 3x - 2.5. So the hypotenuse of the triangle is (3x - 2.5) + (4x - 2.5) which we know is 5x. So we have (3x - 2.5) + (4x - 2.5) = 5x 7x -5 = 5x 2x = 5 x = 2.5 This gives the sides of the rectangle: 3x = 7.5 and 4x = 10. So the area is 75.
I wonder if the puzzle is correct. Diagonal of the rectangle is tangent to all circles, so line segment drawn from center of each circle on that would be perpendicular and would measure to respective radius. Now line segment drawn between yellow and blue is also tangent to them so line segment drawn from their respective center would also be perpendicular and measure to radius forming an impossible rectangle with one side equal to 1.5 (yellow circle radius) while the opposite side measuring 2 (blue circle radius). Am I missing something?
At 2:40, let the angle between the length 4x side and the hypotenuse be 2Θ, then tan(2Θ) = 3x/4x = 3/4. The line drawn from that vertex to the center of the circle bisects that angle. We use the tangent double angle formula (or our notes) to find tan(Θ) = 1/3. The distance from that vertex to the point of tangency on the length 4x side is 4x - 2.5. So, tan(Θ) = (2.5)/(4x - 2.5). Replacing tan(Θ) with 1/3 and solving, x = 5/2. Area of triangle = 6x² = 75/2. Area of rectangle is double the area of the triangle, or 75, as Andy Math also found. In our notes, we should have the tangents of half angles for the 3-4-5 right triangle: if tan(2Θ) = 3/4, the tan(Θ) = 1/3 and if tan(2Θ) = 4/3, the tan(Θ) = 1/2.
I assigned a, b, c, d, and e to various segments. The small triangle was assigned legs a, b, and hypotenuse c. The mid-size triangle was assigned legs b, d, and hypotenuse e. The large triangle was assigned legs c, e, and hypotenuse a+d. Since each pair of segments from the point of tangency on the circle to their vertical intersection are equal and the radius is perpendicular at the point of tangency then the radius, r, can be shown to be r=tan(C/2)*(a+b-c)/2, where angle C is opposite to side c. Right angle C, opposite to hypotenuse c, is 90° in the small triangle and so tan(90°/2)=tan(45°)=√(2)/√(2)=1, then 2r=a+b-c and given that 2r is the diameter of the inscribed circle then 3=a+b-c for the small triangle. Without loss of generality, 4=b+d-e for the mid-size triangle and 5=c+e-(a+d)=c+e-a-d for the large triangle. Summing all equations gives 2b=12, so b=6. Back substitution and applying the Pythagorean theorem gives a=c-3 c^2-(c-3)^2=(c+c-3)(c-c+3)=3(2c-3)=36 2c-3=12 so c=6+1.5=7.5 and d=e-2 e^2-(e-2)^2=(e+e-2)(e-e+2)=2(2e-2)=36 2e-2=18 so e=9+1=10. Having found the rectangle sides c and e then c*e=7.5*10=75 square units.
great video as always! I love how you put up the scale ratio, really got me thinking "oh, maybe he's going to use it to find the other triangle's area" until it dawned on me that the biggest triangle divides the rectangle in half. love your work!
inradius r=(a+b-c)/2; which is 2.5 = (3x+4x-5x)/2 ; so x = 2.5; therefore area of rectangle = 2* area of largest triangle = 2(6)(2.5)^2 = 75 sq.units using inradius is easy too!
Cool, you want a speed race. He's offering content to everybody, in a clear & simple way. Some of you advanced math freaks need to go outside and play, more often.
Bro Pythagorean triplets can be 12,5 and 13 but if we choose them we will get area 93.75. so the technique you used is not correct here by luck you choose correct triplets 3,4 and 5 may be the diameters given help you to choose but mathematically it is not correct way to do so and yes if you checked further by choosing triplets with other diameters given then it is correct
For 30th december Square "?" has a height of 2c+d. Square "12" has a width of 2c + p and a height of c + d - p/2. The triangle a horizontal line creates in the 'turned' square has a height always equal to half it's base - the triangle created is similar to all such triangles, and if you look at the triangle created from cutting in half at the diagonal, the property becomes evident. Since it's a square we have : 2c + p = 12 c + d -p/2 = 12 multiply the second by two : 2c + 2d - p = 24 add to the first one : 4c + 2d +p-p = 36 divide by two : 2c + d = 18 and 2c + d is the height (and thus the width) we were looking for. Hence 18.
"Square "12" has a width of 2c + p and a height of c + d - p/2." Sorry, im dumb. how did you get that the difference between the diagonal of the rotated square (you called it d) and (12-c) is equal to half of p? where did you get that, that's what has me stuck. It looks plausible but I cant see the relation.
@freekingfreaking246 draw in the diagonal d and draw a segment p across the gap between the two purple squares. Doing that creates two congruent isoceles right triangles in the corner of the rotated square; d bisects p so the short side length of each IRT is p/2.
Thank you, I was trying to follow other people's answers with frustration but this made it easy. I kept trying to put sqrt(c) instead of just substituting d for the diagonal of the squares.
An alternate way to solve for the area once we know x - we already know the area of the rectangle is 3x * 4x = 12x, so no need to solve for the area of the larger triangle.
For 12 square, Width is 2s+x, height is s+(s{2}-x/2) Where s is the purple side length and x is the amount from purple to purple. ({2}-1)s=3/2x x=s(2{2}-2)/3 So 12=2s+(2{2}/3)s-(2/3)s=2{2}/3*({2}+1)s > s=36/(2{2}*({2}+1))=18/(2+{2}) For the bigger square, the width is equal to the height, which is 2s+s{2}=s(2+{2})=18
how do you know the difference between the diagonal of the rotated square and (12-s) is equal to half of x? where did you get that, that's what has me stuck. It looks plausible but I cant see the relation.
I'm curious to know how you arrived to the statement that the height is s+(s√2 -x/2), how did you know it turns out to be x/2? Btw I did solve it and with steps and all, and you didn't need to solve for s OR x (especially not x). If hadn't gone after the height of the bigger black square first I wouldn't have noticed at all. Its height is s + s√2 + s = s(2 + √2) The height equation of the 12u square simplified is s(2 + √2) = 18 (after substituting for x/2 = 6 - s of course). Notice anything? Exactly, you can just remove the s(2 + √2) and substitute for the "?" To end up with 18
The bottom corner of the diamond that is cut off by the two squares is an upside-down 45° - 90° - 45° triangle. The base of this triangle is x (the gap between the two squares on the bottom), the height of the upside-down 45° - 90° - 45° triangle is half of its base or x/2. Got it?
@@freekingfreaking246 the first equation I wrote and the equation of x = 12 - 2x, using normal simplification you get: s + (s√2 - (6 - s)) = 12, s + s√2 - 6 + s = 12, s(1 + √2 + 1) = 12 + 6, s(2 + √2) = 18
Job one - - if you start with a rectangle divided up into triangles as shown, HOW DO WE KNOW that the diameters of the other two incircles are 4, and 3, AS CLAIMED? That's what I would want to verify first! The idea that you can cut a rectangle into random triangles, and the incircles go 5,4,3 in nice integers . . . seems sus.
It's not claimed; it's given. Look at the figure. All 3 diameters (3, 4, and 5) are given. I think any Pythagorean triple could have been used as the diameters of those in-circles and created a similar puzzle.
@kenhaley4 I believe what he meant to say was, what makes the author of the question sure they'd line perfectly into a rectangle with those inradii, or if they're aligned perfectly into a rectangle, how does he know the other 2 radii would be acceptable with those numbers.
@@kenhaley4 Why not any 3 side lengths, as long as the 3angle is right? For a triangle with side lengths abc, aren't the incircles necessarily in proportion as a:b:c, in this construction? And if one continues to sub-divide the triangles by drawing altitudes, then the ratios continue. Assume a>b>c. Then the ratios go: a:b:c a:b:c a:b:c and so on, fractally.
Triangle similarity always gave me problems, but my question is, is it obvious that the ratio of the sides of similar triangles is the same as the ratio of the areas of the inscribed circles? I can believe it, but I am thinking about the proof and haven't gotten there.
Answer to the next question: If each side of the three small squares = x The gap between the two square on the right = b 12 = x + (√2)x - b/2 = 2x + b (√2 - 1)x = 3b/2 b = (2√2 - 2)x/3 12 = 2x + (2√2 - 2)x/3 = (4 + 2√2)x/3 x = 18/(2 + √2) = 9(2 - √2) ? = 2x + (√2)x = (2 + √2)(9)(2 - √2) = 18
I continue to struggle with math. However, I still find this interesting to try to follow. Thanks for giving me joy in a subject I never thought I'd have
18, I think (edited since I made a dumb mistake) calling the side of a purple square x, the length we are trying to find y, and the gap between the two purple squares z y=x+x+(root(x^2+x^2)) which simplifies to y=2x+xroot2*** according to the smaller black square horizontally: 2x+z=12 vertically the length is two diagonals minus the overlap the overlap is the diagonal of a square with an inscribed square with sides z touching the centre of each side, which is 2*z which is 2xroot2-2z=12 rearranged makes z=xroot2-6 plug that into the horizontal length to get 2x+xroot2-6=12 2x+xroot2=18 this is the length of the side we need, so y=18
it cant be 12. 12 is the side of the smaller square. how could the bigger square also have side 12 if the bigger square fits more purple squares than the smaller squares?
This is where i think you got wrong: "vertically the diagonal of a purple square minus z is 6, so" If I understand it correctly, you wanna get the diagonal of a forth of the smaller black square, which would be xroot2 + (z/2)root2 [or, alternatively, (x+z/2)root2], but thats not equal to 6, thats equal to 6root2. I also havent been able to find a solution to this one. Ping me if you figure it out
@@freekingfreaking246 I don't know how to ho along your same logic so I'll just add what I did for this part using YOUR variable names. The height of the smaller black square is x + x√2 - m, (m is a tiny part I'll figure out in a minute). Now, see that small area stuck between the 2 squares, I zoomed it and drew it. I drew a parallel line to the bottom as well, touching the pointy stuck end of the square. Now I have an upsidedown right triangle, its upper side is parallel to the horizontal diagonal of its square, which is an isosceles right triangle (45-45-90), hence the small upsidedown triangle is also a (45-45-90) triangle with a base of z. Drawing a perpendicular from its pointy bottom end, it bisects the base and makes 2 congruent triangles. The height of that perpendicular is our m (which I said I'll find a minute ago), the 2 newly formed right triangles are also (45-45-90) triangles, making their base (z/2) equal to their height (m). So our m = z/2 = 6 - x. Zooming back out, we can now say that x + x√2 - (6 - x) = 12, x(2 + √2) = 18. The reason I didn't solve directly for x is because this is easier and faster. The height of the bigger square "y", y = x + x√2 + x = x(2 + √2). Now you get why I didn't solve for x, comparing the 2 equations we get that y = 18
@@freekingfreaking246 thanks for that corrected it, y=18 i think, barring any more stupid mistakes if you draw a purple square in each corner of the 12x12 square and add another square at a 45' angle at the top so the vertical length is equal to twice the diagonal of the purple square minus the overlap the gap between the four squares is a smaller square size z while the overlap is a square with the z square inscribed exactly at the centre of each side. This makes the diagonal of the overlap square 2z, which makes my method simpler since you get a pair of lengths 12=2x+z 12=2xroot2-2z
Let x be the side of the big black square and y the side of the purple square. We have x=2y+y√2. The gap between the two purple squares at the base of the small black square is an isosceles right triangle with a hypotenuse of 12-2y and a height of (12-2y)/2=6-y. Hence, y-(6-y)+y√2=12, and 2y+y√2=18. Therefore, x=18.
I'm not sure its correct 😅😅😅😅 But this is my answer for the puzzle First of all I am gonna be using some terms which are 1.d(Diagonal of small square ) 2.A and a (Side of big square and Side of small Squares respectively) 3. Bigger square at the left is S¹ and bigger square at the right is S² Without further ado lets get to it 🎉 First of all we all know that all sides of a square are equal Now we are asked to find the side of S¹ and we are also given that the size of all small squares are the same meaning that all the small squares are of the same side length Now looking at S² the length of the bottom small square + the length of the diagonal of the top small square all placed inside S² equals to 12 (since all side of a square are equal). We can make up the equation (a + d = 12) and now according to the formula I saw on the internet, the diagonal of a square equals a√2. now substituting this information we now have( a + a √2 = 12) which is (a + 1.4a = 12) giving us (2.4a = 12 )now dividing both sides by 2.4 we get that a = 5.great we are almost done now Okay let's look at S¹ The length of (?) Equals to( a + diagonal + a) which is (a + d + a)giving us( a + a √2 + a) and since we know that a is 5. Substituting we now have (5 + 5√2 + 5)giving us approximately 17 as our answer How exciting?😂
Because the three triangles are similar, their hypotenuses are proportional to the diameters of the circles drawn inside them. If a, b, and c are the hypotenuses of the three triangles, then a/5=b/3=c/4=x, so a=5x, b=3x, and c=4x.
In general, all linear dimensions of similar triangles...including the diameter of the inscribed circles...will be in the same ratio, and use the same scale factor. This is true of any set of similar geometric shapes.
Or, you assigned base 3x and height 4x, so area is 12x^2, so 12(25/4), so 3*25, so 75 units squared. IF, you can assume a circle of 5 means a hypotenuse of 5……..
This guy is gonna spent New Year’s Eve solving like 5 of these puzzles. How exciting!
Let’s put a box around that - that looks important.
Break out the Chinese food!
This will be a fun one.
No fireworks untill the last aggvent video. My 2025 will not begin untill I see all puzzles solved.
Looking back through the "Book of Andy", we had a video (like a year ago) showing the inscribed circle in a 3u-4u-5u triangle has radius 1 (and so area pi). Here, once we arrive at the 3 similar triangles being of side proportions 3-4-5 (@1:47), we can just note the radius of the inscribed circle (say of the largest triangle) to quickly determine the actual dimensions involved. As we have 2.5 units instead of 1, we need to scale a 3u-4u-5u triangle by factor of 2.5 to get the large triangle. Our large triangle must therefore be 7.5u-10u-12.5u, meaning the rectangle is 7.5u x 10u or 75 sq. units.
I solved this slightly differently starting at 2:24, with no further dissection required. You need to draw this to follow, but it's pretty easy. Label the vertices of the triangle A, B, and C starting at the top and going counter-clockwise. Label the points of tangency to the circle A' , B' , and C' with each of these opposite the corresponding vertex of the triangle. Label the center of the circle O.
Now, since OC'BA' is a square and AB = 4x, AC' is (4x - 2.5). We know that the distances from a point outside a circle to the tangent points on the circle are equal. So AB' is also (4x - 2.5). Similarly, CB' and CA' are each = 3x - 2.5. So the hypotenuse of the triangle is (3x - 2.5) + (4x - 2.5) which we know is 5x. So we have
(3x - 2.5) + (4x - 2.5) = 5x
7x -5 = 5x
2x = 5
x = 2.5
This gives the sides of the rectangle: 3x = 7.5 and 4x = 10. So the area is 75.
Really cool
I wonder if the puzzle is correct. Diagonal of the rectangle is tangent to all circles, so line segment drawn from center of each circle on that would be perpendicular and would measure to respective radius. Now line segment drawn between yellow and blue is also tangent to them so line segment drawn from their respective center would also be perpendicular and measure to radius forming an impossible rectangle with one side equal to 1.5 (yellow circle radius) while the opposite side measuring 2 (blue circle radius). Am I missing something?
At 2:40, let the angle between the length 4x side and the hypotenuse be 2Θ, then tan(2Θ) = 3x/4x = 3/4. The line drawn from that vertex to the center of the circle bisects that angle. We use the tangent double angle formula (or our notes) to find tan(Θ) = 1/3. The distance from that vertex to the point of tangency on the length 4x side is 4x - 2.5. So, tan(Θ) = (2.5)/(4x - 2.5). Replacing tan(Θ) with 1/3 and solving, x = 5/2. Area of triangle = 6x² = 75/2. Area of rectangle is double the area of the triangle, or 75, as Andy Math also found.
In our notes, we should have the tangents of half angles for the 3-4-5 right triangle: if tan(2Θ) = 3/4, the tan(Θ) = 1/3 and if tan(2Θ) = 4/3, the tan(Θ) = 1/2.
I assigned a, b, c, d, and e to various segments. The small triangle was assigned legs a, b, and hypotenuse c. The mid-size triangle was assigned legs b, d, and hypotenuse e. The large triangle was assigned legs c, e, and hypotenuse a+d. Since each pair of segments from the point of tangency on the circle to their vertical intersection are equal and the radius is perpendicular at the point of tangency then the radius, r, can be shown to be
r=tan(C/2)*(a+b-c)/2,
where angle C is opposite to side c.
Right angle C, opposite to hypotenuse c, is 90° in the small triangle and so
tan(90°/2)=tan(45°)=√(2)/√(2)=1, then
2r=a+b-c and given that 2r is the diameter of the inscribed circle then
3=a+b-c for the small triangle.
Without loss of generality,
4=b+d-e for the mid-size triangle and
5=c+e-(a+d)=c+e-a-d for the large triangle. Summing all equations gives
2b=12, so b=6.
Back substitution and applying the Pythagorean theorem gives
a=c-3
c^2-(c-3)^2=(c+c-3)(c-c+3)=3(2c-3)=36
2c-3=12 so c=6+1.5=7.5
and
d=e-2
e^2-(e-2)^2=(e+e-2)(e-e+2)=2(2e-2)=36
2e-2=18 so e=9+1=10.
Having found the rectangle sides c and e then
c*e=7.5*10=75 square units.
great video as always! I love how you put up the scale ratio, really got me thinking "oh, maybe he's going to use it to find the other triangle's area" until it dawned on me that the biggest triangle divides the rectangle in half. love your work!
Andy.
We believe in you and we love your content. Youre amazing and you can do this!
inradius r=(a+b-c)/2; which is 2.5 = (3x+4x-5x)/2 ; so x = 2.5; therefore area of rectangle = 2* area of largest triangle = 2(6)(2.5)^2 = 75 sq.units
using inradius is easy too!
He started well but did not seem to know this. Whole thing is solved imediately.
@@RAG981 its not that he didnt know its that he wanted to explain the whole process to us
Cool, you want a speed race. He's offering content to everybody, in a clear & simple way. Some of you advanced math freaks need to go outside and play, more often.
Bro Pythagorean triplets can be 12,5 and 13 but if we choose them we will get area 93.75. so the technique you used is not correct here by luck you choose correct triplets 3,4 and 5 may be the diameters given help you to choose but mathematically it is not correct way to do so and yes if you checked further by choosing triplets with other diameters given then it is correct
For 30th december
Square "?" has a height of 2c+d.
Square "12" has a width of 2c + p and a height of c + d - p/2.
The triangle a horizontal line creates in the 'turned' square has a height always equal to half it's base - the triangle created is similar to all such triangles, and if you look at the triangle created from cutting in half at the diagonal, the property becomes evident.
Since it's a square we have :
2c + p = 12
c + d -p/2 = 12
multiply the second by two : 2c + 2d - p = 24
add to the first one : 4c + 2d +p-p = 36
divide by two : 2c + d = 18 and 2c + d is the height (and thus the width) we were looking for.
Hence 18.
"Square "12" has a width of 2c + p and a height of c + d - p/2."
Sorry, im dumb. how did you get that the difference between the diagonal of the rotated square (you called it d) and (12-c) is equal to half of p? where did you get that, that's what has me stuck. It looks plausible but I cant see the relation.
@freekingfreaking246 draw in the diagonal d and draw a segment p across the gap between the two purple squares. Doing that creates two congruent isoceles right triangles in the corner of the rotated square; d bisects p so the short side length of each IRT is p/2.
Thank you, I was trying to follow other people's answers with frustration but this made it easy. I kept trying to put sqrt(c) instead of just substituting d for the diagonal of the squares.
An alternate way to solve for the area once we know x - we already know the area of the rectangle is 3x * 4x = 12x, so no need to solve for the area of the larger triangle.
the missing length is 18 units
Still seven to do.
Is it even possible?
I'm worried.
How exciting!
I solved the one for tomorrow! First one that I knew how to solve right away and that didn't take me insanely long to do. How exciting!
For 12 square, Width is 2s+x, height is s+(s{2}-x/2)
Where s is the purple side length and x is the amount from purple to purple.
({2}-1)s=3/2x
x=s(2{2}-2)/3
So 12=2s+(2{2}/3)s-(2/3)s=2{2}/3*({2}+1)s > s=36/(2{2}*({2}+1))=18/(2+{2})
For the bigger square, the width is equal to the height, which is 2s+s{2}=s(2+{2})=18
how do you know the difference between the diagonal of the rotated square and (12-s) is equal to half of x? where did you get that, that's what has me stuck. It looks plausible but I cant see the relation.
I'm curious to know how you arrived to the statement that the height is s+(s√2 -x/2), how did you know it turns out to be x/2?
Btw I did solve it and with steps and all, and you didn't need to solve for s OR x (especially not x).
If hadn't gone after the height of the bigger black square first I wouldn't have noticed at all.
Its height is s + s√2 + s = s(2 + √2)
The height equation of the 12u square simplified is s(2 + √2) = 18 (after substituting for x/2 = 6 - s of course).
Notice anything?
Exactly, you can just remove the s(2 + √2) and substitute for the "?" To end up with 18
@@Z-eng0 where did you get 18? like how did you get to this number, other than from other peoples comments?
The bottom corner of the diamond that is cut off by the two squares is an upside-down 45° - 90° - 45° triangle. The base of this triangle is x (the gap between the two squares on the bottom), the height of the upside-down 45° - 90° - 45° triangle is half of its base or x/2. Got it?
@@freekingfreaking246 the first equation I wrote and the equation of x = 12 - 2x, using normal simplification you get: s + (s√2 - (6 - s)) = 12, s + s√2 - 6 + s = 12, s(1 + √2 + 1) = 12 + 6, s(2 + √2) = 18
Job one - - if you start with a rectangle divided up into triangles as shown, HOW DO WE KNOW that the diameters of the other two incircles are 4, and 3, AS CLAIMED? That's what I would want to verify first! The idea that you can cut a rectangle into random triangles, and the incircles go 5,4,3 in nice integers . . . seems sus.
That's a very critical and great question, I'll search it up and see if I can find out the answer to that one.
It's not claimed; it's given. Look at the figure. All 3 diameters (3, 4, and 5) are given. I think any Pythagorean triple could have been used as the diameters of those in-circles and created a similar puzzle.
@kenhaley4 I believe what he meant to say was, what makes the author of the question sure they'd line perfectly into a rectangle with those inradii, or if they're aligned perfectly into a rectangle, how does he know the other 2 radii would be acceptable with those numbers.
OK, had a look at the triangles. The 3 triangles being in 3/4/5 ratio forces the circles to be in 3/4/5 ratio.
@@kenhaley4 Why not any 3 side lengths, as long as the 3angle is right? For a triangle with side lengths abc, aren't the incircles necessarily in proportion as a:b:c, in this construction? And if one continues to sub-divide the triangles by drawing altitudes, then the ratios continue. Assume a>b>c. Then the ratios go:
a:b:c a:b:c a:b:c and so on, fractally.
Triangle similarity always gave me problems, but my question is, is it obvious that the ratio of the sides of similar triangles is the same as the ratio of the areas of the inscribed circles? I can believe it, but I am thinking about the proof and haven't gotten there.
7 puzzles in 1 day (tomorrow), is it possible??
I wonder that too.
He has 2 full days and the rest of today, I think he’ll make it
In which time zone are you? 🤔
There're still 2 days left in 2024, here.
Answer to the next question:
If each side of the three small squares = x
The gap between the two square on the right = b
12 = x + (√2)x - b/2 = 2x + b
(√2 - 1)x = 3b/2
b = (2√2 - 2)x/3
12 = 2x + (2√2 - 2)x/3 = (4 + 2√2)x/3
x = 18/(2 + √2) = 9(2 - √2)
? = 2x + (√2)x = (2 + √2)(9)(2 - √2) = 18
I continue to struggle with math. However, I still find this interesting to try to follow. Thanks for giving me joy in a subject I never thought I'd have
18, I think (edited since I made a dumb mistake)
calling the side of a purple square x, the length we are trying to find y, and the gap between the two purple squares z
y=x+x+(root(x^2+x^2))
which simplifies to y=2x+xroot2***
according to the smaller black square horizontally:
2x+z=12
vertically the length is two diagonals minus the overlap
the overlap is the diagonal of a square with an inscribed square with sides z touching the centre of each side, which is 2*z
which is
2xroot2-2z=12
rearranged makes
z=xroot2-6
plug that into the horizontal length to get
2x+xroot2-6=12
2x+xroot2=18
this is the length of the side we need, so y=18
it cant be 12. 12 is the side of the smaller square. how could the bigger square also have side 12 if the bigger square fits more purple squares than the smaller squares?
This is where i think you got wrong:
"vertically the diagonal of a purple square minus z is 6, so"
If I understand it correctly, you wanna get the diagonal of a forth of the smaller black square, which would be xroot2 + (z/2)root2 [or, alternatively, (x+z/2)root2], but thats not equal to 6, thats equal to 6root2.
I also havent been able to find a solution to this one. Ping me if you figure it out
@@freekingfreaking246 I don't know how to ho along your same logic so I'll just add what I did for this part using YOUR variable names.
The height of the smaller black square is x + x√2 - m, (m is a tiny part I'll figure out in a minute).
Now, see that small area stuck between the 2 squares, I zoomed it and drew it.
I drew a parallel line to the bottom as well, touching the pointy stuck end of the square.
Now I have an upsidedown right triangle, its upper side is parallel to the horizontal diagonal of its square, which is an isosceles right triangle (45-45-90), hence the small upsidedown triangle is also a (45-45-90) triangle with a base of z.
Drawing a perpendicular from its pointy bottom end, it bisects the base and makes 2 congruent triangles.
The height of that perpendicular is our m (which I said I'll find a minute ago), the 2 newly formed right triangles are also (45-45-90) triangles, making their base (z/2) equal to their height (m).
So our m = z/2 = 6 - x.
Zooming back out, we can now say that x + x√2 - (6 - x) = 12, x(2 + √2) = 18.
The reason I didn't solve directly for x is because this is easier and faster.
The height of the bigger square "y", y = x + x√2 + x = x(2 + √2).
Now you get why I didn't solve for x, comparing the 2 equations we get that y = 18
@@freekingfreaking246
thanks for that
corrected it, y=18 i think, barring any more stupid mistakes
if you draw a purple square in each corner of the 12x12 square and add another square at a 45' angle at the top so the vertical length is equal to twice the diagonal of the purple square minus the overlap
the gap between the four squares is a smaller square size z while the overlap is a square with the z square inscribed exactly at the centre of each side. This makes the diagonal of the overlap square 2z, which makes my method simpler since you get a pair of lengths
12=2x+z
12=2xroot2-2z
Let x be the side of the big black square and y the side of the purple square. We have x=2y+y√2. The gap between the two purple squares at the base of the small black square is an isosceles right triangle with a hypotenuse of 12-2y and a height of (12-2y)/2=6-y. Hence, y-(6-y)+y√2=12, and 2y+y√2=18. Therefore, x=18.
Where do you get all these puzzles from? Do you make them up yourself or find them in math competitions?
I'm not sure its correct 😅😅😅😅
But this is my answer for the puzzle
First of all I am gonna be using some terms which are
1.d(Diagonal of small square )
2.A and a (Side of big square and Side of small Squares respectively)
3. Bigger square at the left is S¹ and bigger square at the right is S²
Without further ado lets get to it 🎉
First of all we all know that all sides of a square are equal
Now we are asked to find the side of S¹ and we are also given that the size of all small squares are the same meaning that all the small squares are of the same side length
Now looking at S² the length of the bottom small square + the length of the diagonal of the top small square all placed inside S² equals to 12 (since all side of a square are equal).
We can make up the equation (a + d = 12) and now according to the formula I saw on the internet, the diagonal of a square equals a√2. now substituting this information we now have( a + a √2 = 12) which is (a + 1.4a = 12) giving us (2.4a = 12 )now dividing both sides by 2.4 we get that a = 5.great we are almost done now
Okay let's look at S¹
The length of (?) Equals to( a + diagonal + a) which is (a + d + a)giving us( a + a √2 + a) and since we know that a is 5. Substituting we now have (5 + 5√2 + 5)giving us approximately 17 as our answer
How exciting?😂
why is the scale factor the same as the diameters of the circles?
Because the three triangles are similar, their hypotenuses are proportional to the diameters of the circles drawn inside them. If a, b, and c are the hypotenuses of the three triangles, then a/5=b/3=c/4=x, so a=5x, b=3x, and c=4x.
In general, all linear dimensions of similar triangles...including the diameter of the inscribed circles...will be in the same ratio, and use the same scale factor. This is true of any set of similar geometric shapes.
I believe in you!
Does this mean we're getting like 7 videos in the next 24 hours or so?
How exciting!
Nice puzzle and solution. I did it by equating the exterior tangent segments
bro lives in his own world and i love that. also cool puzzles
Excellent. You deserve way more subscribers than you have!
Or, you assigned base 3x and height 4x, so area is 12x^2, so 12(25/4), so 3*25, so 75 units squared. IF, you can assume a circle of 5 means a hypotenuse of 5……..
The last how exciting was little bit sad
I used the formula for the area S=p*r (p is semiperimeter)
We don't need the diameter value of the biggest circle to solve this problem
Im honestly surprised i managed to atleast solve the first half of the puzzle
day25
a is square side
(12-a)+(6-a)=a·sqrt2
→2a+a·sqrt2=x=18😊
nice one!
Day 25: 18 units
Cool
5÷2=2.5×2.5=6.25×6=37.5
Next problem: 18
Why this is just the devil's magic. Repent you sorcerer.
among us
I'm excited but also nervous since there's so many days left still but only 2 days left now🫨