Once we have xy = 99 and x² + y² = 324, we can set the quadratic equation x⁴ − 324x² + 9801 = 0. Since we would get the same equation with y, the positive solutions 3 √(18 ± √203) are x and y. These can be denested to obtain x = 3/2 (√58 + √14) and y = 3/2 (√58 − √14) in their exact form. PreMaths' amazing solution avoids both the quadratic equation and the denesting operation! 🤩
Wow! I am amazed how you solved it teacher. I was thinking of using Trigonometry, but I kept watching the clip and it is wonderful how you got the answers. Thanks for posting.✨
An alternative solution can be found with tangent secant theorem. Called OH the perpendicular to BC from the center of the bigger circle, we can write: AC : CH = CH : CP 18 : CH = CH : 7 CH = √ 126 Now applying Pythagorean theorem on ABC right triangle we get: y² + (y+ √ 126)² = 18² y² + √ 126y - 99 = 0 then y = (√ 522 - √ 126)/2 x = (√ 522 - √ 126)/2 + √ 126) = (√ 522 + √ 126)/2
Nice! Math is so cool that there are different proof for us to discover. As usual PreMath's solution was more elegant and clever than mine (I brute forced it with a lot of algebra including denesting radicals and rationalizing the denominators).. Your solution is even more concise!
@ 1:06 Thales Theorem is a special case of Euclid's Inscribed Angle Theorem. And of course we immediately engage Pythagorean. Soooooooo cool! 🙂
Spot on!
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You do outstanding work. This one had me stumped.
Excellent!
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At a quick glance, Important to note the lines AP and PC are chords of circles and not the diameters.
@ 5:20 Adding 2xy too equation 2...absolutely ingenious! @ 6:50 , same thing! @ 9:44 , most ingenious! 🙂
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y = R
Tangent secant theorem:
(x -R)² = 7. (7+11)
(x - y)² = 126
x = y + √126
Pytagorean theorem:
x² + y² = (7+11)²
(y + √126)² + y² = 18²
y² + 2.√126.y+126 + y² =324
2y² + 2.√126.y -198 = 0
y² + √126.y -99 = 0
y = 5,811 cm
x = 17,036 cm ( Solved √ )
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Once we have xy = 99 and x² + y² = 324, we can set the quadratic equation x⁴ − 324x² + 9801 = 0.
Since we would get the same equation with y, the positive solutions 3 √(18 ± √203) are x and y.
These can be denested to obtain x = 3/2 (√58 + √14) and y = 3/2 (√58 − √14) in their exact form.
PreMaths' amazing solution avoids both the quadratic equation and the denesting operation! 🤩
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
Nice solution sir
Thanks and welcome
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Wow, brilliant!
Excellent!
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You are awesome. Keep it up 👍
Wow! I am amazed how you solved it teacher. I was thinking of using Trigonometry, but I kept watching the clip and it is wonderful how you got the answers. Thanks for posting.✨
An alternative solution can be found with tangent secant theorem. Called OH the perpendicular to BC from the center of the bigger circle, we can write:
AC : CH = CH : CP
18 : CH = CH : 7
CH = √ 126
Now applying Pythagorean theorem on ABC right triangle we get:
y² + (y+ √ 126)² = 18²
y² + √ 126y - 99 = 0 then
y = (√ 522 - √ 126)/2
x = (√ 522 - √ 126)/2 + √ 126) = (√ 522 + √ 126)/2
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Nice! Math is so cool that there are different proof for us to discover. As usual PreMath's solution was more elegant and clever than mine (I brute forced it with a lot of algebra including denesting radicals and rationalizing the denominators).. Your solution is even more concise!
@@waheisel thank you
Best explanation
Glad you think so!
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Very interesting question sir.
Many many thanks
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Sir, you are a good teacher for us to teach Mathematics. Thank you sir. Our greetings to achieve sir.🎉🎉🎉
Thanks for your continued love and support!
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Love and prayers from the USA! 😀
Beautiful!
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Excellent!
Amazing 👍
Thanks for sharing 😊
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It´s not clear to me why the diagonal passes through the point of tangency, but I believe it.
Sir please tell proof of ceva's theorem and menelau theorem
Please give me time to do my homework. These proofs are little challenging. Cheers
Very nice but in the last we should put the x value in eq. 01
xy=99
17.04*y=99
y=99/17.04
y=5.81...... easy way
❤👍😀
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A Very hard question Sir
Yes! It's a challenging one!
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Beautiful
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Comment for yesterday's Menelaus problem.
Cevians! They're everywhere!
All Hail Giovanni Ceva! 🙂
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Very hard sum
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x°2+y°2=324...x^2=162+√(162^2-99^2)=290,.....mah,i conti non sono interi...ho dei dubbi
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@@PreMath...hai scritto 17.04...allora il mio risultato è ok!!!!
Professor, your source of problems appears to be inexhaustible ❗️