Right Angle In A Quarter Circle

[@mieldavid9433]

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.

[@SQRTime]

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.dominio4765]

💙🙏

[@dickson3725]

Whatt, this one was easy

[@timmylim1844]

Man I’m in eighth grade and googling every word this guy says

[@mieldavid9433]

@@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

[@smylesg]

What makes the first method especially cool is that all the triangles are pythagorean (yes, I said it) triples.

[@SQRTime]

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 😊

[@sabyasachirimpa]

Gogu theorem.

[@ambidextrousseamer7101]

@@sabyasachirimpa 😹😹bc

[@JamesWylde]

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

[@ambidextrousseamer7101]

@@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?

[@peterkwan1448]

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_bkl]

nice one🙂

[@peterkwan1448]

​@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

[@hammer3721]

GoGu

[@nicknicht]

@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 😅

[@smchoi9948]

《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

[@umutklc2182]

There is a simple solution with the idea of A'OC ~ A'DA triangles. First find "r", than find |OC|.

[@yapsiauwsoengie6507]

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.

[@SQRTime]

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_ngl]

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)

[@JLvatron]

I solved by reflection. What's cool about the Cyclic method is that you never actually have to solve for r.

[@TheSourovAqib]

Finally 1 that i can actually do! 😂

[@mathe.dominio4765]

💙🙏

[@Tiqerboy]

Presh titled it as challenging. It wasn't. The real satisfaction comes from solving one of his problems he admits got him stumped.

[@eugene7518]

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.

[@murdock5537]

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...:-)

[@murdock5537]

...and: You don't need Ptolemy theorem: 24 square x 32 square lead to (2r) square...- and via 25 square to "x"...

[@shykitten55]

I would like to be able to remember HALF the tricks you show.

[@advaitkulkarni9634]

nice this problem was a great revision of concepts . 👍🏽👍🏽

[@Waldlaeufer70]

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!

[@SQRTime]

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 😊

[@Waldlaeufer70]

@@SQRTime Thanks for the waypost to your channel!

[@SQRTime]

@@Waldlaeufer70 You are very welcome and thanks for checking.

[@study-2006]

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-sw3gy]

It was an interesting problem. I was able to solve this. I used the first method of reflected quarter circle😌

[@leif1075]

But when youbreflected did you draw the two lines again 24 and 7 or extend the chord like he did??

[@DilipKumar-sw3gy]

I actually reflected the quarter circle and extended the chord.

[@mikefochtman7164]

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.

[@leif1075]

@@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??

[@mikefochtman7164]

@@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.

[@haydenwong659]

Your method are really simple to understand!

[@SQRTime]

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 😊

[@anuradhakhanvilkar3675]

Wow I'm amazed i was able to solve that by myself . Your channel really did make me smarter in just 2 days

[@HackedPC]

I am in 8th grade easily solved it. You don't need to be so proud. It's a very easy problem.

[@natashok4346]

Very good, but in geometry the side of triangle always x>0, not equal with zero

[@jetx_47]

I love how Presh refrains from saying the name of the famed right triangle theorem.

[@SQRTime]

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 😊

[@mathmurthy993]

A beautiful problem on pythogoren triplets mixed with simple concepts

[@mathe.dominio4765]

💙🙏

[@pranshukrishna5105]

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

[@nosystemissafe3507]

Was thinking using some trig ratios

[@priyanshujha144]

I used that extending the quarter cirlce method

[@pintukarmakar3643]

Wonderful ,I did not know Ptolemy theorem superb ❤️❤️

[@asgardianadi772]

Solve for x Presh Talwalker x/x+4 - x/x+7 = 11/30 I think you must made a video on this problem

[@HackedPC]

no real solution

[@kenhaley4]

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.dominio4765]

🙏💙

[@archturusdeydas3989]

bruh lol ok

[@efi3825]

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.

[@rohangeorge712]

i like this problem i solved it using the first problem, i know nothing about ptolemty theorem

[@charlesdang2557]

A delightful problem with pythagorean triplets. Thanks, buddy

[@SQRTime]

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 😊

[@facilitateclub3690]

Very interesting problem

[@050138]

Yay! Solved mentally in 2 minutes! 🙂

[@abdallajama4832]

Which class take 🤔

[@050138]

@@abdallajama4832 master class take....

[@user-yg97f5hfvh]

That hypotenuse 25 forces me to write down 15

[@pankaj71]

I thought of the first method and used it to easily solve this 🔥

[@dilipkumarpatel481]

Big fan bro 🤜🤛

[@spelunkerd]

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!

[@jackhandma1011]

Utilizing cyclic quadrilaterals is the cool one.

[@JarppaGuru]

2:53 yes you smart. i told it first comment xD

[@mr.maccaman2]

Was it raining when you recorded this?

[@jercki72]

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!

[@jercki72]

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.

[@HackedPC]

@@jercki72 hey I'm in 8th grade too and solved it easily.

[@codyx8273]

2nd one ive gotten right

[@pinkystinkeye5120]

I love these videos!

[@SQRTime]

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 😊

[@MDExplainsx86]

How could you prove that DC || CA'

[@muneebmuhamed43]

i am in grade 8 and i managed to understand this.

[@hamzasenouci5452]

15

[@davidhernandez9985]

Permission to come aboard, the first-time viewer!

[@SQRTime]

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 😊

[@Starteller]

Try this question: "What was the temperature(s) of the film inside the camera while being on the Moon?"

[@benoitschwob3498]

Wow, amazing job... :-o

[@SQRTime]

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 😊

[@dwaipayandattaroy9801]

24

[@abhijeetsarker5285]

So Beautiful 😍

[@akraoglory2515]

Best 🤠

[@TylerDURDEN935]

İ love you bro

[@tariqislam4915]

😂😂😂

[@jeffrey5952]

Guessed the answer

[@SQRTime]

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 😊

[@mathsx5887]

The right angle was kind of pushing me to use the first method.

[@SQRTime]

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 😊

[@mitchellclark4377]

I found the reflection method to be quite elegant; well done. And all integer values too!

[@hughcaldwell1034]

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.

[@SAMax2268]

I actually used Ptolemys Theorem to solve this…as my first approach 😁

[@SQRTime]

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 😊

[@leif1075]

Why..who even knoes that theorem..don't you find it unintuitive??

[@Musfiqur_anik]

Me too

[@davidhernandez9985]

Now that a logical solution to solve this mathematical problem.

[@efi3825]

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.

[@Skandalos]

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.

[@ayanahmedkhan2580]

Can you plz explain it to me

[@GamingZone-hv3wq]

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

[@deerh2o]

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!

[@jeremiahlyleseditor437]

The Ptolemy Solution escaped me.

[@Wmann]

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.

[@mirmahdimurtazaomi2719]

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!?

[@MaxEng1492]

100% because he created the problem that way

[@williamangelogonzales148]

Is this a repost? I think I remember this problem in this channel but not in night mode.☺️

[@SQRTime]

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 😊

[@philipkudrna5643]

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.

[@SQRTime]

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 😊

[@HackedPC]

Good job

[@yashindersingh5148]

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_bkl]

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)))

[@SQRTime]

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 😊

[@kushalshedha4891]

I am very happy becauseI today I learnt a new theorem that is Ptolemy's theorem

[@waffles6132]

If we had the angle (AĆO) WE could use the scalar product and find AO then with Pythagorean theorem find OC mesure.

[@rssl5500]

Amazing I solved it with the first method !!!! :D

[@saadkhondoker1921]

I used Ptolemy to solve it but was searching for the 1st solution. Thanks for showing both.

[@quigonkenny]

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.

[@zdrastvutye]

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

[@abhishekpatil1063]

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.

[@daapdary]

At 2:43, "Let's focus on the triangle AOC." Hey, keep politics out of it! 😀

[@JarppaGuru]

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

[@elooouan]

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

[@evanj3535]

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.

[@divijsharma5610]

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 !!

[@rachelanderson9318]

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

[@SQRTime]

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-sy9tj]

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.

[@sea34101]

I used a third approach that I found pretty convoluted... method #1 is neat.

[@SQRTime]

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-oeyd4020]

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

[@JarppaGuru]

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

[@HackedPC]

I did exactly the same way as the solution 1. But my rusted brain used algebra to find the answer😂😂

[@YuvrajSingh-kr1vy]

Happy holi to all 😀😀

[@FirstnameismyLastname]

I just went straight towards the upscaled 3-4-5 right triangle after I got the hypotenuse of AOC

[@JarppaGuru]

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

[@JarppaGuru]

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_2007]

Please Come Back To White Background! Love that!🤍

[@pragya857]

How do I find puzzle videos for 7 year old in your channel? Do you have any seperate playlist for kids puzzle?

[@b77vedantmore51]

Can anyone please suggest that ,which app/software is MYD use to make YT videos

[@poiuyqwert6545]

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 .

[@RicoGrafity]

2:31 i need help with this part...why not using 25²+25²=(2r)² to find r...

[@harikatragadda]

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

[@SQRTime]

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 😊

[@roddurde5462]

Is it me just ,an Asian , or some kids used pythagorus 3, 4, 5 rule to find OC?

[@not_vinkami]

idk if any of you guys believe but I calculated the right answer in mind without any support or hint

[@m.guypirate6900]

How did you know that DC intersected A' in the first problem?

[@vidyadharjoshi5714]

AC = 25. Extend by 25 to meet extended AO at E. So AO = 20. OC = sqrt(25sq - 20sq) = 15

[@JarppaGuru]

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

[@leif1075]

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..

[@WaiWai-qv4wv]

Good