This is why I love this channel. At the start I thought It is impossible for me to solve this. But when the solution was presented, it is intuitive enough for me to understand.
@SQRTime2 жыл бұрын
Hi Miel. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@mathe.dominio47652 жыл бұрын
💙🙏
@dickson37252 жыл бұрын
Whatt, this one was easy
@timmylim18442 жыл бұрын
Man I’m in eighth grade and googling every word this guy says
@mieldavid94332 жыл бұрын
@@timmylim1844 me too, I'm entering tenth grade so some of his problems are out of my league. But this holds true to geometry problems that I can understand
@smylesg2 жыл бұрын
What makes the first method especially cool is that all the triangles are pythagorean (yes, I said it) triples.
@SQRTime2 жыл бұрын
Hi there . If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@sabyasachirimpa2 жыл бұрын
Gogu theorem.
@ambidextrousseamer71012 жыл бұрын
@@sabyasachirimpa 😹😹bc
@JamesWylde2 жыл бұрын
I bet it would drive Presh nuts to learn that not only is it Pythagoras Theorem, but that's just a special case of the British Flag Theorem. en.m.wikipedia.org/wiki/British_flag_theorem
@ambidextrousseamer71012 жыл бұрын
@@JamesWylde woah! It's a cool theorum, thanks for mentioning it. But that theorum cann't be proven without applying the Pythagoras theorum itself. So if you say that Pythagoras theorum is a special case of British Flag Theorum, it is contradictory. And if you say that this question is a special case of BFT, please care to explain me. Where did to you take point P, and where is the rectangle?
@peterkwan14482 жыл бұрын
I used the laws of cosine to solve the equation. By Pythagoras's Theorem, AC = 25. Let x = ∠AOD. By the laws of cosine for triangle AOD: 2r²(1-cos x)=24² => cos x=1-(24²/2r²) => cos² x= 1 - (24²/r²) + (12²24²/r⁴) => sin² x= (24²/r⁴) (r²-144) => sin x = (24/r²) √(r²-144) By the laws of cosine for triangle OCD: 25² - 2 r √(25²-r²) sin x = 7² => 12 r = √[(25²-r²)(r²-144)] => r⁴ - 25²r² + 25²*12² = 0 By quadratic equation solution: r² = 400 or r²=225 If r²=225, cos x = -0.28, which means x>90°. Therefore r² = 400, and OC = √(25²-400) = 15.
@lexus_bkl2 жыл бұрын
nice one🙂
@peterkwan14482 жыл бұрын
@Insert Username OD² + OC² - 2 OD OC cos (90-x) = BC² => r² + (25²-r²) - 2 r √(25²-r²) sin x = 7² => 25² - 2r √(25²-r²) sin x = 7²
@hammer37212 жыл бұрын
GoGu
@nicknicht2 жыл бұрын
@Insert Username for me it was very intuitive to continue DC. I wouldn’t call it reflecting, but making a more familiar picture. Sometimes you just get an urge to draw some elements for aesthetic reasons 😅
@smchoi99482 жыл бұрын
《Please note: "O" & "x" used in the host's video are accordingly replaced by O' & L: O' is not the origin used below.》 Align the whole figure w/ the Cartesian plane, so that (1) A is at the origin (0,0); & (2) AO' lies on +ve x-axis. Let C₀ be the circle completed from the arc portion of the given quarter-sector, & its radius r. C₀'s centre is then at O'(r,0). The equation of the line containing AD is y = mx for some m>0, so D is at (k,mk) for some k>0. As D is on C₀, (k-r)² + (mk)² = r² ...(i) => m² = 2r/k - 1 ...(i'). As |AC|² = |AD|² + |DC|², |AC| = √(24² + 7²) = 25 ...(*). Let |O'C| = L. As |AO'| = r, by (*), r² + L² = |AC|² = 25² ...(ii). As C is at (r,L), |CD| = 7 => (k-r)² + (mk-L)² = 7² ...(iii). As |AD| = 24, k² + (mk)² = 24² => m² = 24²/k² - 1 ...(iv). (iv) -> (i'): 24²/k² - 1 = 2r/k - 1 r = 288/k ...(#) (#) -> (ii): (288/k)² + L² = 25² L² = 25² - 288²/k² ...(¥) (i) - (iii): (mk)² - (mk-L)² = r² - 7² 2mkL - L² = r² - 7² 2mkL + 7² = r² + L² 2mkL + 7² = 25² (by (ii)) mkL = 288 ...($) (iv) & (¥) -> ($): [√(24²/k² - 1)](k)[√(25² - 288²/k²)] = 288 25²(k²)² - (24)²(25²)k² + (24²)(288²) = 0 k² = 2(25±7)(12/5)² k = 96/5 or k = 72/5 ...(£) (£) -> (#): If k = 96/5, r = 288/(96/5) = 15 (rej. as kk). (£) -> (¥): L² = 25² - 288²/(72/5)² = 225 => L = 15
@umutklc21822 жыл бұрын
There is a simple solution with the idea of A'OC ~ A'DA triangles. First find "r", than find |OC|.
@yapsiauwsoengie65072 жыл бұрын
Half of AC is center point of smaller circle that will be pass through AOCD. Perpendicular line from O to AD will devide AD into two same parts and crossing the center point of the small circle so we easily got it's value. Next calculate the quarter circle's radius and then OC.
@SQRTime2 жыл бұрын
Hi Yap. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@nice_mf_ngl2 жыл бұрын
Presh, please try Indian mathematics Olympiad problems too, just in case you want the names first this was the system of examinations (till 2018) - PRMO (pre regional mathematics Olympiad) - stage 1 RMO ( regional mathematics Olympiad ) INMO ( Indian national mathematics Olympiad ) Since then this new system has been followed (due to COVID it was changed ) - IOQM - part A ( Indian Olympiad qualifier of mathematics ) IOQM - part B (basically just INMO)
@JLvatron2 жыл бұрын
I solved by reflection. What's cool about the Cyclic method is that you never actually have to solve for r.
@TheSourovAqib2 жыл бұрын
Finally 1 that i can actually do! 😂
@mathe.dominio47652 жыл бұрын
💙🙏
@Tiqerboy2 жыл бұрын
Presh titled it as challenging. It wasn't. The real satisfaction comes from solving one of his problems he admits got him stumped.
@eugene75182 жыл бұрын
China had no known tradition of building lifelike sculptures before Qin's reign. According to Li Xiu Zhen a senior archeologist at the Terra Cotta army site this departure of scale and style likely occurred when when influences arrived in China specifically from ancient Greece, i.e. chariots, hairstyles. They believe the appearance may have been modeled/inspired by Greek sculptures. Hence Chinese and Greeks had contact then and they exchanged knowledge such as the Pythagorean theorem.
@murdock55372 жыл бұрын
Amazing! When you find 7 + 25 = 32, you solved the problem. :-) Many thanks! By the way: We have 3 Pythagorean triples (ACD, AOC, AA'D (2 x 3-4-5, 1 x 7-24-25). That's why we'v got integers...:-)
@murdock55372 жыл бұрын
...and: You don't need Ptolemy theorem: 24 square x 32 square lead to (2r) square...- and via 25 square to "x"...
@shykitten55 Жыл бұрын
I would like to be able to remember HALF the tricks you show.
@advaitkulkarni96342 жыл бұрын
nice this problem was a great revision of concepts . 👍🏽👍🏽
@Waldlaeufer702 жыл бұрын
1) Draw a line from A to C. You'll get a right triangle. So, according to Pythagoras, AC = 25 2) Complete the Thales' circle to the right and mark the end opposite A as E. 3) According to symmetry, AC equals CE. So, DE = 7 + AC = 32 4) Calculate the diameter of the Thales' circle: d² = 24² + 32² => d = 40 5) Calculate OC using the right triangle ACO: OC² = AC² - (0.5d)² = 25² - 20² = 225 => OC = 15 Now, let's watch the video... P.S. I have never heard of Ptolemy's Theorem before. Nice solution!
@SQRTime2 жыл бұрын
Hi Wald. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@Waldlaeufer702 жыл бұрын
@@SQRTime Thanks for the waypost to your channel!
@SQRTime2 жыл бұрын
@@Waldlaeufer70 You are very welcome and thanks for checking.
@study-20062 жыл бұрын
we can also draw a line from o to d that perpendicularly bisects ad and then by using cosine and some trigonometry we can get x just another answer ..
@DilipKumar-sw3gy2 жыл бұрын
It was an interesting problem. I was able to solve this. I used the first method of reflected quarter circle😌
@leif10752 жыл бұрын
But when youbreflected did you draw the two lines again 24 and 7 or extend the chord like he did??
@DilipKumar-sw3gy2 жыл бұрын
I actually reflected the quarter circle and extended the chord.
@mikefochtman71642 жыл бұрын
When I saw a 90 degree angle at point D, first thing I thought of was that extending DC must intersect the circle directly opposite A on the diameter (what he labeled A' ). But beyond that, I was a bit lost. When he actually drew the reflected quarter circle, only then did I see the isosceles triangle and it became clear.
@leif10752 жыл бұрын
@@mikefochtman7164 why would you think of extending the 7 line instesd of reflecting both tje quarter circle AND the two lines if length 24 and 7..that seems more intuitive to me since more symmetry??
@mikefochtman71642 жыл бұрын
@@leif1075 I suppose it could be thought that way. I just noticed a 90 degree angle and remembered the double-angle theorem, so the extended line 7 would intersect the circle at 180 degrees, a full diameter. Just occurred to me before the idea of reflecting. Just what came to mind first.
@haydenwong6592 жыл бұрын
Your method are really simple to understand!
@SQRTime2 жыл бұрын
Hi Hayden. If you are interested in math competitions, please consider kzbin.info/www/bejne/f6GbpXtqoLCJkLs and other videos in the Olympiad playlist. Hope you enjoy 😊
@anuradhakhanvilkar36752 жыл бұрын
Wow I'm amazed i was able to solve that by myself . Your channel really did make me smarter in just 2 days
@HackedPC2 жыл бұрын
I am in 8th grade easily solved it. You don't need to be so proud. It's a very easy problem.
@natashok43462 жыл бұрын
Very good, but in geometry the side of triangle always x>0, not equal with zero
@jetx_472 жыл бұрын
I love how Presh refrains from saying the name of the famed right triangle theorem.
@SQRTime2 жыл бұрын
Hi Jetx. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@mathmurthy9932 жыл бұрын
A beautiful problem on pythogoren triplets mixed with simple concepts
@mathe.dominio47652 жыл бұрын
💙🙏
@pranshukrishna51052 жыл бұрын
What about joining mid point of hypotenuse to O in the cyclic quadrilateral solution from there also we can find AC instead of applying plotemy's theorm
@nosystemissafe35072 жыл бұрын
Was thinking using some trig ratios
@priyanshujha1442 жыл бұрын
I used that extending the quarter cirlce method
@pintukarmakar36432 жыл бұрын
Wonderful ,I did not know Ptolemy theorem superb ❤️❤️
@asgardianadi7722 жыл бұрын
Solve for x Presh Talwalker x/x+4 - x/x+7 = 11/30 I think you must made a video on this problem
@HackedPC2 жыл бұрын
no real solution
@kenhaley42 жыл бұрын
A fun part of this one was that all the important line lengths turned out to be integers. I still find it sad that Presh refuses to say "Pythagorean" even though that's the common name of the theorem at the heart of this puzzle--no matter which way you solve it. It doesn't matter that it wasn't discovered by Pythagorus--that's still what it's called. The Fibonacci sequence wasn't discovered by Fibonacci, but we still call it that.
@mathe.dominio47652 жыл бұрын
🙏💙
@archturusdeydas39892 жыл бұрын
bruh lol ok
@efi38252 жыл бұрын
At least he tried to be sensitive about it. But it might be more valuable to have proper naming conventions. It helps making learning easier.
@rohangeorge7122 жыл бұрын
i like this problem i solved it using the first problem, i know nothing about ptolemty theorem
@charlesdang25572 жыл бұрын
A delightful problem with pythagorean triplets. Thanks, buddy
@SQRTime2 жыл бұрын
Hi Charles. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@facilitateclub36902 жыл бұрын
Very interesting problem
@0501382 жыл бұрын
Yay! Solved mentally in 2 minutes! 🙂
@abdallajama48322 жыл бұрын
Which class take 🤔
@0501382 жыл бұрын
@@abdallajama4832 master class take....
@user-yg97f5hfvh2 жыл бұрын
That hypotenuse 25 forces me to write down 15
@pankaj712 жыл бұрын
I thought of the first method and used it to easily solve this 🔥
@dilipkumarpatel4812 жыл бұрын
Big fan bro 🤜🤛
@spelunkerd2 жыл бұрын
Ha ha. Again, I found a way to make this more complicated than it had to be. I used Thales theorem, followed by the law of cosines, then solved three equations with three unknowns. Doh!
@jackhandma10112 жыл бұрын
Utilizing cyclic quadrilaterals is the cool one.
@JarppaGuru2 жыл бұрын
2:53 yes you smart. i told it first comment xD
@mr.maccaman22 жыл бұрын
Was it raining when you recorded this?
@jercki722 жыл бұрын
Alright let's go Let C_1 be the circle which quarter is drawn in the problem. Applying Pythagoreas in the right triangle DAC => AC = 25. Let I be the midpoint of [AC], AI = IC = 25/2 Since DAC is right triangle, its circumcenter is I. Let (d) be the perpendicular bisector of [AD]. Let C_2 be the circumcircle of DAC, its center is I. A, D ∈ C_2, thus I ∈ (d). A, D ∈ C_1 which center is O, thus O ∈ (d) OAC is a right triangle, so its circumcenter is the midpoint of the hypotenuse [AC], which is I. In fact, the circumcircle of OAC is also C_2. Thus O ∈ C_2, thus OI = AI = 25/2. Let J be the midpoint of [AD]. J ∈ (d). Triangles DAC and JAI are similar with factor 1/2. Thus IJ = CD/2 = 7/2. Thus OJ = OI +I J = 16. Applying Pythagoreas in the right triangle JAO => AO = 20. Applying Pythagoreas in the right triangle OAC, finally OC = 15. For some reason, geometry problems like these are a lot harder for me, I feel like we didn't do much of it past 8th grade unfortunately. In this case I took a long time trying to figure out the radius of C_1, and ended up trying to make the construction from scratch, starting from segment [AD], which eventually gave me insight as of where circle C_1 comes from. Still took me a long time. Also, now that I think about it, point B was never used both in my reconstruction and in the solution. And I have to appreciate as an intention from the problem maker that all the lengths where Pythagoreas is needed end up being integers, truly amazing!
@jercki722 жыл бұрын
Ok, so I actually spent some time trying to figure out a trick like in solution 1, and I am actually kind of mad I couldn't find it and make it work ;-; And never heard of cyclic quadrilaterals, unlucky.
@HackedPC2 жыл бұрын
@@jercki72 hey I'm in 8th grade too and solved it easily.
@codyx82732 жыл бұрын
2nd one ive gotten right
@pinkystinkeye51202 жыл бұрын
I love these videos!
@SQRTime2 жыл бұрын
Hi Pinky ☺️. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@MDExplainsx865 ай бұрын
How could you prove that DC || CA'
@muneebmuhamed432 жыл бұрын
i am in grade 8 and i managed to understand this.
@hamzasenouci54522 жыл бұрын
15
@davidhernandez99852 жыл бұрын
Permission to come aboard, the first-time viewer!
@SQRTime2 жыл бұрын
Hi David. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@Starteller2 жыл бұрын
Try this question: "What was the temperature(s) of the film inside the camera while being on the Moon?"
@benoitschwob34982 жыл бұрын
Wow, amazing job... :-o
@SQRTime2 жыл бұрын
Hi Benoit. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@dwaipayandattaroy98012 жыл бұрын
24
@abhijeetsarker52852 жыл бұрын
So Beautiful 😍
@akraoglory25152 жыл бұрын
Best 🤠
@TylerDURDEN9352 жыл бұрын
İ love you bro
@tariqislam49152 жыл бұрын
😂😂😂
@jeffrey59522 жыл бұрын
Guessed the answer
@SQRTime2 жыл бұрын
Hi Jeffrey. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@mathsx58872 жыл бұрын
The right angle was kind of pushing me to use the first method.
@SQRTime2 жыл бұрын
Hi Florian. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@mitchellclark43772 жыл бұрын
I found the reflection method to be quite elegant; well done. And all integer values too!
@hughcaldwell10342 жыл бұрын
Yeah, finding the reflection was very satisfying, though I'm annoyed I missed the cyclic method. I don't think he mentioned in the video, but not only is the 15-20-25 a scaled up 3-4-5, but so is the 24-32-40, which I thought was fun.
@SAMax22682 жыл бұрын
I actually used Ptolemys Theorem to solve this…as my first approach 😁
@SQRTime2 жыл бұрын
Hi Sam. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@leif10752 жыл бұрын
Why..who even knoes that theorem..don't you find it unintuitive??
@Musfiqur_anik2 жыл бұрын
Me too
@davidhernandez99852 жыл бұрын
Now that a logical solution to solve this mathematical problem.
@efi38252 жыл бұрын
It's a little sad that you rarely get to use Ptolemys Theorem in these geometry puzzles, even though it's a more powerful version of the Pythagorean Theorem.
@Skandalos2 жыл бұрын
I always try to construct the situation in geogebra to understand the constraints and find possible solution paths. Helped me nicely once again. I ignored the quarter circle, just constructed the right triangle with legs 24 and 7. Then I used Thales theorem and constructed a semi circle above the hypotenuse because point O is a right angle corner. Then I realized that point O had to be on a perpendicular bisector of the 24 leg since OAD is isoceles with the radius of the quarter circle as both legs. With the help of the intercept theorem it was easy to determine the radius of the quarter circle and then the length of OC. Also all three right triangles involved contained pythagorean triplets. Nice one.
@ayanahmedkhan25802 жыл бұрын
Can you plz explain it to me
@GamingZone-hv3wq Жыл бұрын
Could have been done like this also probably... Right angle with legs 7 and 24, giving out hypotenuse as 25, following if we gaze the figure properly BC will make up the radius if added 4 times(intuition), therefore BC is 1/4 of r, i.e. r/4, eventually, In right-angled triangle AOC, 25²=r²+(3r/4)² Giving out r as 20, substituting value gives CO is 15
@deerh2o2 жыл бұрын
Presh's first method (and the way I solved it) goes from a 7- 24-25 triple to two different scaled up 3-4-5 triples. How fun!
@jeremiahlyleseditor4372 жыл бұрын
The Ptolemy Solution escaped me.
@Wmann2 жыл бұрын
I actually got it right, but not in a way that I would hope to get right in a test if I were to use the way I used. I first found AC with the Pythagorean Theorem which is 25, and guessed that the triangle AOC is a 3, 4 and 5 right triangle. First, I set OC as 3 Then, I knew that 25 is 5 multiplied by 5, so 3 must be multiplied by 5 too. And whadaya know, 3 multiplied by 5 is 15. Cool.
@mirmahdimurtazaomi27192 жыл бұрын
The fact that all the values we got are integers is so cool..... I mean what's the possibility of you square rooting some numbers that are derived from previous square rooted number and you get an integer!?
@MaxEng14922 жыл бұрын
100% because he created the problem that way
@williamangelogonzales1482 жыл бұрын
Is this a repost? I think I remember this problem in this channel but not in night mode.☺️
@SQRTime2 жыл бұрын
Hi William. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@philipkudrna56432 жыл бұрын
Before watching. If you extend the quarter circle to the other side, you get an inscribed right triangle. The missing piece of the sidelength is 25 (24^2+7^2), which makes the large inscribed triangle 24-32-40. (So the diameter is 40 or the radius 20.) Now back to the actual problem, the height. Again a right triangle: 25^2-20^2=225. So the height is 15.
@SQRTime2 жыл бұрын
Hi Philip. If you are interested in math competitions, please consider kzbin.info/www/bejne/f6GbpXtqoLCJkLs and other videos in the Olympiad playlist. Hope you enjoy 😊
@HackedPC2 жыл бұрын
Good job
@yashindersingh51482 жыл бұрын
Ayo presh! Please notice this🙏 In right angled triangle ADC, AC={(24^2)+(7^2)} AC=25 and In right angled triangle AOC AC={(AO^2)+(OC^2)} 25={2x^2} [Let AO=OC=x] Then, x= [25(2^1/2)]/2 So, CO should be 17.6.
@lexus_bkl2 жыл бұрын
General solution: Let one of the length be 'p' & one of the length be 'b'. Then, OC = sqrt(1/2 (p^2 + b^2 - b sqrt(p^2 + b^2)))
@SQRTime2 жыл бұрын
Hi Lexus. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@kushalshedha48912 жыл бұрын
I am very happy becauseI today I learnt a new theorem that is Ptolemy's theorem
@waffles61322 жыл бұрын
If we had the angle (AĆO) WE could use the scalar product and find AO then with Pythagorean theorem find OC mesure.
@rssl55002 жыл бұрын
Amazing I solved it with the first method !!!! :D
@saadkhondoker19212 жыл бұрын
I used Ptolemy to solve it but was searching for the 1st solution. Thanks for showing both.
@quigonkenny6 ай бұрын
Connect A and C. Perform Pythagoras on ∆CDA to find AC. 7² + 24² = AC² AC² = 49 + 576 = 625 AC = √625 = 25 Reflect the quarter circle to the right about OB to create a semicircle with E, where AE is the diameter of the semicircle. By Thales's Theorem, if a right triangle is inscribed into a circle, the hypotenuse forms the diameter. Thus we see that EC is a reflection of AC about OB, and ED is a straight line segment of length 7+25 = 32. As DA = 24, we find that ∆EDA is an 8:1 ratio 3-4-5 right triangle, and this AE = 5(8) = 40. As AE is a diameter of the semicircle, the radius OA = r = 20. As we now know that OA = 20 and CA = 25, we again find a 3-4-5 right triangle in ∆AOC, this one at a 5:1 ratio, and so OC = 3(5) = 15.
@zdrastvutye2 жыл бұрын
es war nicht ganz leicht, ausführen mit amstrad locomotive basic: 10CLS:WINDOW OPEN:WINDOW FULL:nu=25:DIM x(3),y(3),xk(nu),yk(nu) 20nui=15:t2=0:REM anzahl der schritte in der grafik 30lr=10:lk1=24:lk2=7:lh=SQR(lk1^2+lk2^2):flab=lk2/lk1:sw=.1 40x1=-lr:y1=0:x2=sw-lr:xe=x2:x(0)=0:y(0)=0:y(3)=0:GOSUB 50:GOTO 120 50x2=xe:y2=SQR(lr^2-x2^2):dy=y1-y2:dx=x1-x2:lku1=SQR(dx*dx+dy*dy) 60dxk=-flab*dy:dyk=flab*dx:lku2=SQR(dxk^2+dyk^2):REM STOP 70xek=x2+dxk:yek=y2+dyk:dl=xek/lr:? xek;yek;dl:REM gosub 30 80lhu=SQR(lku1^2+lku2^2):IF lhu=0 THEN 100 90lrr=lr*lh/lhu 100x(1)=x2:y(1)=y2:x(2)=xek:y(2)=yek:x(3)=x(2) 110RETURN 120xe1=xe:dl1=dl:xe=xe+sw:IF ye>10*lh THEN STOP 130xe2=xe:GOSUB 50:IF dl1*dl>0 THEN 120 140xe=(xe1+xe2)/2:GOSUB 50:IF dl*dl1>0 THEN xe1=xe ELSE xe2=xe 150IF ABS(dl)>1e-8 THEN 140 ELSE CLOSE WINDOW 4 160PRINT xe,ye,yek:fv=lk1/lku1:PRINT lhu*fv;lr*fv:GOSUB 370 170FOR a=0 TO nu:wa=PI/2*a/nu:wa=wa+PI/2:xk(a)=lr*COS(wa):yk(a)=lr*SIN(wa) 180NEXT a:lhu2=SQR(lr^2+yek^2):PRINT lhu2:GOSUB 370 190PRINT "Der gesuchte Radius ist=";lrr:GOSUB 220:co=1:t2=1 200xeb=xe:FOR b=0 TO nui:dxb=(-lr-xeb)*b/nui:xe=dxb+xeb:GOSUB 50 210GOSUB 240:NEXT b:END 220x(0)=x1:y(0)=y1:x(1)=x2:y(1)=y2:x(2)=xek:y(2)=yek:x(3)=0:y(3)=0 230mass=400:stu=1:co=1:REM die befehle trennen fuer goto/sub *** 240FOR a=0 TO 3:ia=a+1:IF a=3 THEN ia=0 250xmin=x(0):ymin=y(0):GOTO 270 260xb=x-xmin:yb=y-ymin:xb=xb*mass:yb=yb*mass:RETURN 270x=x(a):y=y(a):GOSUB 260:xb1=xb:yb1=yb: 280x=x(ia):y=y(ia):GOSUB 260:xb2=xb:yb2=yb:GOTO 300 290LINE xb1;yb1,xb2;yb2 COLOUR co STYLE stu:RETURN 300GOSUB 290:NEXT a:IF t2=1 THEN 370 310co=4:FOR a=0 TO nu-1:ia=a+1:x=xk(a):y=yk(a) 320GOSUB 260:xb1=xb:yb1=yb:x=xk(ia):y=yk(ia):GOSUB 260 330xb2=xb:yb2=yb:GOSUB 290:NEXT a: 340x=x(2):y=y(2):GOSUB 260 350xp=ROUND(xek*fv):yp=ROUND(yek*fv) 360CIRCLE xb;yb,40 COLOUR 4:MOVE xb;yb:REM PRINT COLOUR (4);xp;yp 370a$=INKEY$:IF a$="" THEN 370 ELSE RETURN
@abhishekpatil10632 жыл бұрын
JoinAC and DO. We get AC=25 by pythagoras thm. Then Sine (Angle DAC)=7/25. So Angle DAC= 16.26 degrees. So Angle ACD=90-16.26=73.7 degrees. Now since AngleADC + AngleAOC=90+90=180 degrees. Hence AOCD is cyclic. So Angle AOD=AngleACD=73.7 degrees. Similarly AngleDAC=AngleDOC=16.26 degrees. Now AO=OD So Angle DAO=AngleADO=(180-73.7)/2=53.12 degrees. So AngleODC=90-53.12=36.88 degrees. Now using sine rule in triangleOCD, Sine(36.88)/x= Sine(16.26)/7 So x= 15 units.
@daapdary2 жыл бұрын
At 2:43, "Let's focus on the triangle AOC." Hey, keep politics out of it! 😀
@JarppaGuru2 жыл бұрын
0:39 why OC? when we not know OA we can calculate sqrt(24*24+7*7)=AC sqrt(AC*AC-OA*OA)=OC sqrt(AC*AC-OC*OC)=OA what is radius? LOL there is no angles OC change if radius change that initial triangle just turn but keep same size. can we calculate it LOL based on what? we not know angle or radius its same problem as how far is closest star we not know size to get distance we not know distance to get size and light speed nobody measure. and if we compare 2 images and see star moved based on other we can calculate....how we know how far and big is others we compare it LOL we not have those lol. get it? they are so wrong everything
@elooouan2 жыл бұрын
my dumbass did 625-400=115 instead on 225 which would result in sqrt(125)=11.18 instead of sqrt(225)=15, legit made a mental error on the very last step
@evanj35352 жыл бұрын
I solved it another way. Angle ACD has a tangent of 24/25 = 0.96, so it is about 73.74 degrees. Angle ACD + Angle ACA' = 180 degrees, and Angle ACD + Angle CAO + Angle CA'O = 180, so Angle ACA' = Angle CAO + Angle CA'O. Angles CAO and CA'O are base angles of isoceles triangle ACA', so they are equal. Therefore Angle CAO is half of Angle ACA', which is about 36.87 degrees. The sine of that angle = 0.6, and that = opposite/hypotenuse, which means opposite/25 = 0.6, so opposite = 15.
@divijsharma56102 жыл бұрын
Used cyclic quadrileteral logic , bcoze it seemed obvious as angle sum of two oppsite angles were 180 = (90 + 90) (otherwise this question can not be solved) , I do not remember the threorem : Ptolemys theorem. So I used important property to solve it : angles substented by an arc to any point of circumference are same. Used Sine and cosine theorem inside it. I am happy , I do not remember any theorem , just solved it by basics. But the first : mirror image solution is OP. Thanks man !!
@rachelanderson93182 жыл бұрын
Cyclic Quadrilateral ≡ Sum of the opposite angles are all 180 degrees ⇒ Ptolemy's theorem: Sum of products of the opposite sides = Product of two diagonals
@SQRTime2 жыл бұрын
Hi Rachel. If you are interested in math competitions, please consider kzbin.info/www/bejne/f6GbpXtqoLCJkLs and other videos in the Olympiad playlist. Hope you enjoy and have a wonderful time 😊
@RenatoSilva-sy9tj2 жыл бұрын
It could be solved just completing the circle and using Pytaghoras for the triangle rectangle hypotenuse 2r and sides 24 and 25+7, then r=20. So the other rectangle triangle hypotenuse r and sides x and 25, then x=15.
@sea341012 жыл бұрын
I used a third approach that I found pretty convoluted... method #1 is neat.
@SQRTime2 жыл бұрын
Hi Sea. If you are interested in math competitions, please consider kzbin.info/www/bejne/j2a0amenh9-rac0 and other videos in the Olympiad playlist (geometry/algebra). Hope you enjoy 😊
@zoey-oey-oeyd40202 жыл бұрын
same! i figured out that it was a cyclic quadrilateral and used the property of cyclic quadrilaterals that the angle between a side and a diagonal is equal to the angle between the opposite side and the other diagonal (as well as the fact that AOD is an isosceles triangle) to do a bunch of angle algebra and trig and eventually ended up with 15
@JarppaGuru2 жыл бұрын
real question and answer is how we know initial triangle imaginery line is match for another side of arc.bcoz you can dran triangle IN this size ARC and it wont fit, but still IF you samu it is so you get same result. what will be wrong. this maths work if it so, but it might be not. you can draw that triangle different angle same size radius. so how you know before calculate it is so or is it not. what are changes it is or it is not LOL another you with trianle in arc. one is you expand r so it fit. so how you know lol example. IF r is same and you turn triangle so 24is like 23 and 7 would be 7+ ofc it would not fit on your calculations. that imaginery expanded line would not go another side i can see it and i allready cam it. SO HOW YOU KNOW it is so? you got result but you not know is it correct. yes it is if you try find that correct angle this happen, but it could also be it should not fit lol
@HackedPC2 жыл бұрын
I did exactly the same way as the solution 1. But my rusted brain used algebra to find the answer😂😂
@YuvrajSingh-kr1vy2 жыл бұрын
Happy holi to all 😀😀
@FirstnameismyLastname2 жыл бұрын
I just went straight towards the upscaled 3-4-5 right triangle after I got the hypotenuse of AOC
@JarppaGuru2 жыл бұрын
2:36 why so hard its sqrt(32*32+24*24) yes its same thing. we not need 2r we allready know it will be half lol
@JarppaGuru2 жыл бұрын
2:14 yes yes that easy IF that arc is size that meet that imaginery ANOTHER 24 triangle can be different angle so it not meet lol but thats not question we create it so it is so. i was more like it have to fit in this ARC we not know. we just make arc size we need LOL. yes that easy
@RS_20072 жыл бұрын
Please Come Back To White Background! Love that!🤍
@pragya8572 жыл бұрын
How do I find puzzle videos for 7 year old in your channel? Do you have any seperate playlist for kids puzzle?
@b77vedantmore512 жыл бұрын
Can anyone please suggest that ,which app/software is MYD use to make YT videos
@poiuyqwert65452 жыл бұрын
I am Moaaz the first solution for me to solve it that we can draw a cyclic quadrateral from the mid point of AC then there are four directions of the radius of 12.5 then we get angle DAC then use Alkashi law to get OC ... And the mid point of AD .
@RicoGrafity2 жыл бұрын
2:31 i need help with this part...why not using 25²+25²=(2r)² to find r...
@harikatragadda2 жыл бұрын
Can also be done without solving for r : Drop a perpendicular line DE on OA. The altitude of a Right triangle with sides 24 and 32 on the hypotenuse = DE = 24*32/√(24² + 32²) Triangle DEA' is Similar to Triangle COA' CO = DE*25/32 = 15
@SQRTime2 жыл бұрын
Hi Hari. If you are interested in math competitions, please consider kzbin.info/www/bejne/f6GbpXtqoLCJkLs and other videos in the Olympiad playlist. Hope you enjoy 😊
@roddurde54622 жыл бұрын
Is it me just ,an Asian , or some kids used pythagorus 3, 4, 5 rule to find OC?
@not_vinkami2 жыл бұрын
idk if any of you guys believe but I calculated the right answer in mind without any support or hint
@m.guypirate69002 жыл бұрын
How did you know that DC intersected A' in the first problem?
@vidyadharjoshi5714 Жыл бұрын
AC = 25. Extend by 25 to meet extended AO at E. So AO = 20. OC = sqrt(25sq - 20sq) = 15
@JarppaGuru2 жыл бұрын
1:31 yes yes but that triangle can be different angle and that not happen. in this case bcoz its reversed we allready know answer
@leif10752 жыл бұрын
Dodnt everyone else think when he reflected the quarter corcle he would reflect the two line segments 24 and 7 instead of jjst extending the line..that would work too..