...interesting to see different approaches. Another simple one: draw lines from centre of circle to 60 degree vertex and to tangent point on vertical to form Isosc. triangle with angles = 45 degrees and hypotenuse = sqrt(2); then 'a = sqrt(2)[sin (15)], so ' b^2 = 1 - 2[(sin (15))^2] = cos(30)', and '2b = 2[sqrt{cos(30)}]'.
This was a fun one! Speaking of root 3 minus 1 over 2, how about a 3D geometry puzzle: What's the radius of a cylinder of height 1that's inscribed 3D diagonally and symmetrically within a unit cube?
Thanks, SyberMath; I was lucky to stumble across this elegant problem. Here's a synopsis of my approach (I need to leave stuff out, space, but comment if you want). The circle is centered at (1,1) so it's (x-1)^2+(y-1)^2=1. The slant of the hypotenuse of a 30-60-90 triangle with y-intercept = 2 is y=2-(x/root3). If we set the ys equal we eventually get x=(3+sqrt(3) +- 4throot(108))/4. (4throot(108) is root(6*root(3)).) Both these x values are needed as they are for each of the two points. We need to find the ys to match. Substitute this in to the original equation as x to find y=2-(x/root3)=(7root(3)-3 +- 4throot(108))/(4root3). The chord length is the root of the sum of the squares of the differences of the dimensional values; for x, 4throot(27/4) and for y, 4throot(3/4). So the chord length = root((root(3/4))+(root(27/4))) = root2 * 4throot3. This took our buddy about 12 minutes here, subtracting intro and such. Me, it took 12 hours perhaps, and over a half hour to type this. But it was FUN. That's what matters.
Easy one. Just take centre of circle as (1,1) and 2 sides of triangle are on axes. Now you can easily find equation of hypotenuse. Now calculate perpendicular distance from (1,1) to the line(hypotenuse).after that you can easily find chord length
Yayy!!! Solved correctly using a much simpler method, which uses a known theorem.... If A is a point outside a circle, a tangent drawn from A to the circle touches the circle at B, and another straight line from A interests the circle in two points C and D, forming chord CD, then AB*AB = AC*AD
A very nice task, however, the solution is heavily incomplete. You have voluntarily assigned the shorter leg of the rectangular triangle to be equal to 2 (just check the written formulation of the problem!). You must consider two more cases: the former is when the longer leg is equal to 2, and the latter is when the height erected from the right corner is equal to 2. Surprisingly, all three cases have the solution, each. And the answers are different; thus, the problem has three answers what makes it even more interesting! :)
Alternate Solution( much shorter): The area of the triangle is 2SQRT3. Connect the center of the circle to the vertices of the triangle. Three other triangles are formed. The sum of the areas of these three triangles is equal to the area of the original triangle. Find a, then find b.
My way was by getting the erea of the 30-90-60 triangle by two methode to get the perpendicular to the cord and after that were done !!(by pythagorean theorem )
@@SyberMath if you connect from the center of the circle to the vertisis of the big right triangle you get three triangle inside of the right triangle From there you have one unknown (the perpendicular to the cord from the center) from there we can get the area of the right triangle in two way (one by adding the three triangles areas and one by b*h/2) We can get the base and all we need by (trigonometry) then you get a linear equation in terme of the only unknown and finaly i use phytagreon theorm like what you did . (I think phytagreon theorem is one of the bigest discovery in geometric history )
@@SyberMath why cant you draw two triangles connecting the center of the circle to the endpoints of thebchord and then you have an isosceles triangle and two unknowns..the corner angles theta and the length of the chord..two unknowns so you need two equations..you have the law of sines and the law of cosines..2 equationsright there....why didnt you try this..there is absolutely no logical reason why this shouldnt work..yet it doesnt seem to..didnt you try thst first??
...interesting to see different approaches. Another simple one: draw lines from centre of circle to 60 degree vertex and to tangent point on vertical to form Isosc. triangle with angles = 45 degrees and hypotenuse = sqrt(2); then 'a = sqrt(2)[sin (15)],
so ' b^2 = 1 - 2[(sin (15))^2] = cos(30)', and '2b = 2[sqrt{cos(30)}]'.
This was a fun one!
Speaking of root 3 minus 1 over 2, how about a 3D geometry puzzle: What's the radius of a cylinder of height 1that's inscribed 3D diagonally and symmetrically within a unit cube?
That would be pretty interesting!
Really good puzzle ! Loved it ! And thanks to @Qermaq , the original poster of this problem in comments section !
Great 👍!
Thanks pal. I was lucky to stumble upon this.
@@Qermaq Absolutely! Are you ready for a radical journey? 😁
@@SyberMath Heh. My kryptonite!
Thanks, SyberMath; I was lucky to stumble across this elegant problem. Here's a synopsis of my approach (I need to leave stuff out, space, but comment if you want).
The circle is centered at (1,1) so it's (x-1)^2+(y-1)^2=1. The slant of the hypotenuse of a 30-60-90 triangle with y-intercept = 2 is y=2-(x/root3).
If we set the ys equal we eventually get x=(3+sqrt(3) +- 4throot(108))/4.
(4throot(108) is root(6*root(3)).)
Both these x values are needed as they are for each of the two points. We need to find the ys to match.
Substitute this in to the original equation as x to find y=2-(x/root3)=(7root(3)-3 +- 4throot(108))/(4root3).
The chord length is the root of the sum of the squares of the differences of the dimensional values; for x, 4throot(27/4) and for y, 4throot(3/4).
So the chord length = root((root(3/4))+(root(27/4))) = root2 * 4throot3.
This took our buddy about 12 minutes here, subtracting intro and such. Me, it took 12 hours perhaps, and over a half hour to type this. But it was FUN. That's what matters.
Nice! Thanks again for the suggestion and sharing your approach! You're exactly right! It's the journey that matters not just the destination!
Easy one. Just take centre of circle as (1,1) and 2 sides of triangle are on axes. Now you can easily find equation of hypotenuse. Now calculate perpendicular distance from (1,1) to the line(hypotenuse).after that you can easily find chord length
Good!
Care to elaborate your solution? I also tried this solution but got stuck halfway
Yayy!!! Solved correctly using a much simpler method, which uses a known theorem....
If A is a point outside a circle, a tangent drawn from A to the circle touches the circle at B, and another straight line from A interests the circle in two points C and D, forming chord CD, then
AB*AB = AC*AD
Nice!
A very nice task, however, the solution is heavily incomplete. You have voluntarily assigned the shorter leg of the rectangular triangle to be equal to 2 (just check the written formulation of the problem!). You must consider two more cases: the former is when the longer leg is equal to 2, and the latter is when the height erected from the right corner is equal to 2. Surprisingly, all three cases have the solution, each. And the answers are different; thus, the problem has three answers what makes it even more interesting! :)
From a sphere with radius r is derived the largest regular triangular pyramid. What are its dimensions? Thank you! ...
Alternate Solution( much shorter):
The area of the triangle is 2SQRT3. Connect the center of the circle to the vertices of the triangle. Three other triangles are formed. The sum of the areas of these three triangles is equal to the area of the original triangle. Find a, then find b.
Very good!
How would you get the length of the chord from that?
Thanks for the nice problem, SyberMath :)
Any time!
Can I post a problem on divisors ? Which has the potential to make it into one of your videos ?
Sure! Why not? I hope it's not a very hard one 😁
@@SyberMath Suppose that the positive divisors of a positive integer n are 1 = d1 < d2 < d3 < ... < dk = n where k >= 5 . Given that k
My way was by getting the erea of the 30-90-60 triangle by two methode to get the perpendicular to the cord and after that were done !!(by pythagorean theorem )
Can you elaborate on that?
@@SyberMath if you connect from the center of the circle to the vertisis of the big right triangle you get three triangle inside of the right triangle
From there you have one unknown (the perpendicular to the cord from the center) from there we can get the area of the right triangle in two way (one by adding the three triangles areas and one by b*h/2)
We can get the base and all we need by (trigonometry) then you get a linear equation in terme of the only unknown and finaly i use phytagreon theorm like what you did .
(I think phytagreon theorem is one of the bigest discovery in geometric history )
@@tonyhaddad1394 good thinking
@@BCS-IshtiyakAhmadKhan thank u !!!
Wow, nicely done. Thank you sir.
Glad you liked it!
@@SyberMath why cant you draw two triangles connecting the center of the circle to the endpoints of thebchord and then you have an isosceles triangle and two unknowns..the corner angles theta and the length of the chord..two unknowns so you need two equations..you have the law of sines and the law of cosines..2 equationsright there....why didnt you try this..there is absolutely no logical reason why this shouldnt work..yet it doesnt seem to..didnt you try thst first??
excellent video
Thank you very much! 😊
That's a cool problem!
height is 2, but which height of the triangle?
The one that is perpendicular to the base 😁
I was confused like you
@@SyberMath how are you supposed to decide which one's the base and which one's the height as they are pretty much interchangeable here
It wouldn't matter which was the height or base, the problem would not change.
3:57 hahahahahah
😂
Thank you very mach......
You are welcome!
Nice problem good job !!!!!
Thank you! Cheers!
@@SyberMath cheerss 🍻🍻🍻
I solved it using coordinate geometry and it was really satisfying!
Nice!
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