The result does not depend on the value of the circle radius. For r = 0, we have: AO²+OC² = BO²+DO² 33²+r²+x²+r²=85²+r²+35²+r² 33²+x²=85²+35² José Osorio from Brazil.
I agree Jose. Since the radius of the circle was not given therefore the problem should work with all values of R. It this is not the case the problem is indeterminate.
DA =a + b, AB =c + d (separation at the center of the circle), r - radius of the circle #1 a^2 + c^2 - r^2 = 35^2 #2 b^2 + c^2 - r^2 = 33^2 #3 b^2 + d^2 - r^2 = 85^2 a^2 + d^2 - r^2 = #1 - #2 + #3 = x^2 => x =sqrt( 35^2-33^2+85^2 )
@@gojo8063In his derivation, a and b are the vertical distances from the bottom to the centre of the circle, and from the centre to the top. c and d are the horizontal distances from the left of the rectangle to the centre, and from the centre to the right. Each of his equations has triangles involving horizontal, vertical and diagonals on the left (four quadrants), and tangents, radii and diagonals on the right. Draw everything on a diagram and you'll see.
British Flag Theorem is certainly used here (or derived firstly). That would use the four corners joined to the circle centre. But you then have to take the additional step of showing the BFT also applies to the tangent lengths.
@@kevinmorgan2317 This exercise is requesting for X value, not for any demonstration. Besides of that, Radius of circle is not given, so it induce us to think that the result doesn't depend on that radius. And yes, the BFT theorem also applies to the tangent lengths. Anybody wants to demonstrate it??? despite is not required. This video obtained the same result, using a different method, so this video is that demonstration !!!
this is the 4th time i have written a code for this but yet i don't know why some numbers for r=... won't lead to a solution: 10 dim x(3),y(3):r=20:l1=35:l3=85:l4=33:sw=.1:xm=r*1.3+sw:goto 40 20 ym=sqr(l1^2+r^2-xm^2):lh=sqr(l4^2+r^2-xm^2)+ym 30 lb=sqr(l3^2+r^2-(ym-lh)^2)+xm:return 40 gosub 20 50 if lh0 then 130 150 xs1=(xs11+xs12)/2:gosub 100:if dg1*dg>0 then xs11=xs1 else xs12=xs1 160 if abs(dg)>1E-10 then 150:rem den thalessatz anwenden 170 xmt=(xm+lb)/2:ymt=ym/2:rt=sqr((xm-xmt)^2+(ym-ymt)^2) 180 xs2=xm-r:goto 220 190 ys2=ym-sqr(r^2-(xs2-xm)^2):l2=sqr((xs2-lb)^2+ys2^2) 200 dgu1=(xs2-xmt)^2/l1^2:dgu2=(ys2-ymt)^2/l1^2:dgu3=rt^2/l1^2: 210 dg=dgu1+dgu2-dgu3: return 220 gosub 190 230 xs21=xs2:dg1=dg:xs2=xs2+sw:if xs2>10*lb then stop 240 xs22=xs2:gosub 190:if dg1*dg>0 then 230 250 xs2=(xs21+xs22)/2:gosub 190:if dg1*dg>0 then xs21=xs2 else xs22=xs2 260 if abs(dg)>1E-10 then 250 else print "der gesuchte abstand=";l2 270 xs3=xm:goto 310 280 disy=l3^2-(xs3-lb)^2:if disy0 then 330 350 xs3=(xs31+xs32)/2:gosub 280:if dg1*dg>0 then xs31=xs3 else xs32=xs3 360 if abs(dg)>1E-10 then 350 370 xmt=xm/2:ymt=(lh+ym)/2: xs4=xm-r:rt=sqr(xmt^2+(ymt-lh)^2):goto 410 380 disy=rt^2-(xs4-xmt)^2 :if disy0 then 430 450 xs4=(xs41+xs42)/2:gosub 380:if dg1*dg>0 then xs41=xs4 else xs42=xs4 460 if abs(dg)>1E-10 then 450 470 x(0)=0:y(0)=0:x(1)=lb:y(1)=0:x(2)=x(1):y(2)=lh:x(3)=0:y(3)=y(2):mass=1E3/(l1+l2+l3+l4)*4 480 goto 500 490 xbu=x*mass:ybu=y*mass:return 500 xba=0:yba=0:for a=1 to 4:ia=a:if ia=4 then ia=0 510 x=x(ia):y=y(ia):gosub 490:xbn=xbu:ybn=ybu:goto 530 520 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 530 gosub 520:next a:gcol 4:x=xm:y=ym:gosub 490:circle xbu,ybu,r*mass :gcol 8 540 xba=0:yba=0:x=xs1:y=ys1:gosub 490:xbn=xbu:ybn=ybu:gosub 520 550 gcol7:x=lb:y=0:gosub 490:xba=xbu:yba=ybu:x=xs2:y=ys2:gosub 490:xbn=xbu:ybn=ybu:gosub 520 560 gcol8: x=lb:y=lh:gosub 490:xba=xbu:yba=ybu:x=xs3:y=ys3:gosub 490:xbn=xbu:ybn=ybu:gosub 520 570 x=0:y=lh:gosub 490:xba=xbu:yba=ybu:x=xs4:y=ys4:gosub 490:xbn=xbu:ybn=ybu:gosub 520 108.666398% 59.142812 xm=26.1 ym=30.7211653 der gesuchte abstand=85.7962703 > run in bbc basic sdl and hit ctrl tab to copy from the results window
It has a shortcut on that given any length Suppose you has a rectangle ABCD or square with circle inside givem that the origin is at any point inside the triangle with unknown radius. Then each verteces of the rectangle was connected with a of tangency thru the circle. Given that there was a missing value of line C (from the vertex of the triangle to the tabgent point of the circle. The solution is always: A²+C²=B²+D² Example A=9 B=x C=4 D=1 So using the formula B= 4 times sqrt of 6 Is this right? Kindly validate please
I really enjoyed this - and so many of your other puzzle problems. You have a special knack for choosing interesting challenges for your audience (like this one!) and your solutions are ingenious.: we are all lucky to have found you here. Thank you for brightening my day!
I took one look at this and realized immediately the answer must be 87. No matter the radius of the circle or where it is placed within the rectangle, the lengths of the corners A and C tangents will always equal the lengths of the corners B and D tangents. A + C = B + D 85 + 35 = 120 thus: 120 - 33 = 87 being the length of X (the tangent of corner C) The overly complicated methodology of Math Booster gave the incorrect answer of √7361 which is 85.79627032 Squaring the tangent lengths will not give the correct answer, that is of use only if you are looking for the length of each corner ABCD to the centre of the circle, e.g the hypotenuse of the right angle triangles in the video, but you would also need to know the radius of the circle.
you are wrong though not him, his answer is correct, you got it wrong by saying A + C = B + D while in fact the right thing to say would be A^2 + C^2 = B^2 + D^2 which is just the british flag theorem if you select the radius of the circle to be 0, since the radius can be any number, 0 is also a valid choice.
ahm....from the drawn you are assuming that u can make a 90 degree angle line to the center of the circle...but that isn't "evident" on the draw, but i guess it's not possible to solve if that is not the case.
The problem says they're *tangent* lines to the circle. That means each one is, er, flat against a point on the circle and at right angles to the radius at that point.
Seramalatuhgn enkuwan yanten x lifelglh qerto righthnm alfelg deffar azza eski leqeqegnna braseh zabbiya bicha nurrr ynenen x ante aydelehm mitnegregn kante shi etf ymibeltu ngrewgnalna begeza ejh atqlel bayhon yanten x let me tell you yante x yaahya sega alga silut amed ytebalew is best fit for you ok hule khusr ywurre tilq yelewm ytebalew haq new tilq millionair bileh litakebrew sitl bymnderu ltelkasha were endahyaw sega silkesekes setagegnew yane new ekkhkhkjjhkhkhkhjgeeeey miyasegnhh ene badaf endashagn tefeqdolnal ekhkjkjkhkhk allah zerzuryahn ezaaae allah yargewna ameen😮😮😮
Just a little too tedious, along with the not-so-easy-on-the-ear English pronunciation - the video could be almost half its length without losing everything, and not every note written must also be narrated. The solution is nice and beautiful - but I think you could also give 10-20 seconds before solving, on the way you looked at the problem, and how you attacked it, and show the viewers your "war plan", not just the technical details narrated so indifferently - that one KNOWS you had the solution complete to start with. Give it some life! let children not only see how you solve, but develop an appetite for STRUGGLING with such questions. Thank you anyway.
run the following in bbcbasic and change the number for "r=" in line 20: 10 print "brain station end result will shock you":dim x(1,3),y(1,3) 20 l1=33:l3=85:l4=35:r=25:la=60:sw=la/47:sw=2*pi/59:print la:rem @zoom%=1.4*@zoom% 30 s1=sqr(l1^2+r^2):s3=sqr(l3^2+r^2):s4=sqr(l4^2+r^2):print s1,s3,s4:goto 50 40 lhu=(d1^2-d2^2+d3^2)/2/d3:hu=sqr(d1^2-lhu^2):return 50 d1=s1:d2=s4:d3=la:gosub 40:xm=hu:print "xm=";xm:disy=r*r+l1^2-xm^2:if disy0 then w1=w else w2=w 180 if abs(dw)>1E-10 then 170 else return 230 w=sw:xu=0:yu=0:gosub 110:x(1,0)=xs:y(1,0)=ys:w=w+sw:xu=lb:gosub 110:x(1,1)=xs:y(1,1)=ys 240 w=w+sw:yu=la:gosub 110:x(1,2)=xs:y(1,2)=ys:w=w+sw:xu=0:gosub 110:x(1,3)=xs:y(1,3)=ys 250 x(0,0)=0:y(0,0)=0:x(0,1)=lb:y(0,1)=0:x(0,2)=lb:y(0,2)=la:x(0,3)=0:y(0,3)=la 260 xmin=0:xmax=lb:ymin=xm-r:ymax=xm+r 270 masx=1200/lb:masy=850/la:if masx
The result does not depend on the value of the circle radius.
For r = 0, we have:
AO²+OC² = BO²+DO²
33²+r²+x²+r²=85²+r²+35²+r²
33²+x²=85²+35²
José Osorio from Brazil.
I agree Jose. Since the radius of the circle was not given therefore the problem should work with all values of R. It this is not the case the problem is indeterminate.
Brillant solution !!
Green formula + Pink formula ---> AO¢2 +CO¢2 = BO¢2 +DO¢2
I think that the most key part for me is to think more, and then to get pink formula imformation. Good thought!
😮😮
DA =a + b, AB =c + d (separation at the center of the circle), r - radius of the circle
#1 a^2 + c^2 - r^2 = 35^2
#2 b^2 + c^2 - r^2 = 33^2
#3 b^2 + d^2 - r^2 = 85^2
a^2 + d^2 - r^2 = #1 - #2 + #3 = x^2
=> x =sqrt( 35^2-33^2+85^2 )
Please explain with diagram
Most efficient method so far!
@@gojo8063In his derivation, a and b are the vertical distances from the bottom to the centre of the circle, and from the centre to the top. c and d are the horizontal distances from the left of the rectangle to the centre, and from the centre to the right.
Each of his equations has triangles involving horizontal, vertical and diagonals on the left (four quadrants), and tangents, radii and diagonals on the right. Draw everything on a diagram and you'll see.
Find X? --- X is in the lower right in the diagram. Easy!
Sure. That was easy ;-)
X=? Not to find where is X 😅
wow what a imaginatiin plz reply😂😂
Thank u
Sure you never went to school😅@informatikasmkmusanganjuk8540
R=0
British Flag theorem:
x²+33²=35²+85²
x= 85,796 cm ( Solved √ )
British Flag Theorem is certainly used here (or derived firstly). That would use the four corners joined to the circle centre. But you then have to take the additional step of showing the BFT also applies to the tangent lengths.
@@kevinmorgan2317
This exercise is requesting for X value, not for any demonstration.
Besides of that, Radius of circle is not given, so it induce us to think that the result doesn't depend on that radius.
And yes, the BFT theorem also applies to the tangent lengths.
Anybody wants to
demonstrate it??? despite is not required.
This video obtained the same result, using a different method, so this video is that demonstration !!!
this is the 4th time i have written a code for this but yet
i don't know why some numbers for r=... won't lead to a solution:
10 dim x(3),y(3):r=20:l1=35:l3=85:l4=33:sw=.1:xm=r*1.3+sw:goto 40
20 ym=sqr(l1^2+r^2-xm^2):lh=sqr(l4^2+r^2-xm^2)+ym
30 lb=sqr(l3^2+r^2-(ym-lh)^2)+xm:return
40 gosub 20
50 if lh0 then 130
150 xs1=(xs11+xs12)/2:gosub 100:if dg1*dg>0 then xs11=xs1 else xs12=xs1
160 if abs(dg)>1E-10 then 150:rem den thalessatz anwenden
170 xmt=(xm+lb)/2:ymt=ym/2:rt=sqr((xm-xmt)^2+(ym-ymt)^2)
180 xs2=xm-r:goto 220
190 ys2=ym-sqr(r^2-(xs2-xm)^2):l2=sqr((xs2-lb)^2+ys2^2)
200 dgu1=(xs2-xmt)^2/l1^2:dgu2=(ys2-ymt)^2/l1^2:dgu3=rt^2/l1^2:
210 dg=dgu1+dgu2-dgu3: return
220 gosub 190
230 xs21=xs2:dg1=dg:xs2=xs2+sw:if xs2>10*lb then stop
240 xs22=xs2:gosub 190:if dg1*dg>0 then 230
250 xs2=(xs21+xs22)/2:gosub 190:if dg1*dg>0 then xs21=xs2 else xs22=xs2
260 if abs(dg)>1E-10 then 250 else print "der gesuchte abstand=";l2
270 xs3=xm:goto 310
280 disy=l3^2-(xs3-lb)^2:if disy0 then 330
350 xs3=(xs31+xs32)/2:gosub 280:if dg1*dg>0 then xs31=xs3 else xs32=xs3
360 if abs(dg)>1E-10 then 350
370 xmt=xm/2:ymt=(lh+ym)/2: xs4=xm-r:rt=sqr(xmt^2+(ymt-lh)^2):goto 410
380 disy=rt^2-(xs4-xmt)^2 :if disy0 then 430
450 xs4=(xs41+xs42)/2:gosub 380:if dg1*dg>0 then xs41=xs4 else xs42=xs4
460 if abs(dg)>1E-10 then 450
470 x(0)=0:y(0)=0:x(1)=lb:y(1)=0:x(2)=x(1):y(2)=lh:x(3)=0:y(3)=y(2):mass=1E3/(l1+l2+l3+l4)*4
480 goto 500
490 xbu=x*mass:ybu=y*mass:return
500 xba=0:yba=0:for a=1 to 4:ia=a:if ia=4 then ia=0
510 x=x(ia):y=y(ia):gosub 490:xbn=xbu:ybn=ybu:goto 530
520 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
530 gosub 520:next a:gcol 4:x=xm:y=ym:gosub 490:circle xbu,ybu,r*mass :gcol 8
540 xba=0:yba=0:x=xs1:y=ys1:gosub 490:xbn=xbu:ybn=ybu:gosub 520
550 gcol7:x=lb:y=0:gosub 490:xba=xbu:yba=ybu:x=xs2:y=ys2:gosub 490:xbn=xbu:ybn=ybu:gosub 520
560 gcol8: x=lb:y=lh:gosub 490:xba=xbu:yba=ybu:x=xs3:y=ys3:gosub 490:xbn=xbu:ybn=ybu:gosub 520
570 x=0:y=lh:gosub 490:xba=xbu:yba=ybu:x=xs4:y=ys4:gosub 490:xbn=xbu:ybn=ybu:gosub 520
108.666398% 59.142812
xm=26.1 ym=30.7211653
der gesuchte abstand=85.7962703
>
run in bbc basic sdl and hit ctrl tab to copy from the results window
It has a shortcut on that given any length
Suppose you has a rectangle ABCD or square with circle inside givem that the origin is at any point inside the triangle with unknown radius. Then each verteces of the rectangle was connected with a of tangency thru the circle. Given that there was a missing value of line C (from the vertex of the triangle to the tabgent point of the circle.
The solution is always:
A²+C²=B²+D²
Example
A=9
B=x
C=4
D=1
So using the formula
B= 4 times sqrt of 6
Is this right? Kindly validate please
I have the same theory, i havent proof it tho
I appreciate your teaching style
Love from 🇱🇰
I really enjoyed this - and so many of your other puzzle problems. You have a special knack for choosing interesting challenges for your audience (like this one!) and your solutions are ingenious.: we are all lucky to have found you here. Thank you for brightening my day!
I think I need subtitles. By the way, what kind of genius must you be to solve this?! Maximum respect from me.
This is another great one. Heck. It was worth it just to learn the 5² rule.
One can quickly see that this problem will have the same solution regardless of the size of the circle.
How do you think, what country has best geometry scholary. Maybe Hungary , Austria
Leaving your fraction with the square root of 2 is not solving this problem.
Practically Marlen Theorem
оце я розумію "завдання". 👍
hello russian
Nice application and derivation of the British Flag Theorem.
I think we can also use trigonometric ratio
Ex: x/ sin × : y /sin y
😊
But, the operation is not as easy as we think. We need a trigononetric table ...but maybe it will be faster..
Ups sorry...I mean algorithms table
so 85....basically....got it in like 2 seconds.
Greetings from Poland
That is almost 85.8!
Que questão bonita. Vale que: AO^2 + CO^2 = BO^2 + DO^2.
This problem is very very beautiful.
Darimana kamu bisa mendapatkan semua soal ini 🤔
Mes félicitations pour ce travail remarquable
Bhi 16minat mea q hal hoga short turm mea bato
Thanks for your job
So similar to Turkey NMO 1st Round
2023. See tubitak.gov
13:10 23^2 (?)
Bravo! Well done.
35+85=33+x
x=87
شما استاد بزرگی در ریاضی هستی❤
Thanks!
Thank you for supporting the channel. It means a lot for me 😊
It's the best program.
asnwer=95 isit
I took one look at this and realized immediately the answer must be 87. No matter the radius of the circle or where it is placed within the rectangle, the lengths of the corners A and C tangents will always equal the lengths of the corners B and D tangents.
A + C = B + D 85 + 35 = 120 thus:
120 - 33 = 87 being the length of X (the tangent of corner C)
The overly complicated methodology of Math Booster gave the incorrect answer of √7361 which is 85.79627032
Squaring the tangent lengths will not give the correct answer, that is of use only if you are looking for the length of each corner ABCD to the centre of the circle, e.g the hypotenuse of the right angle triangles in the video, but you would also need to know the radius of the circle.
you are wrong though not him, his answer is correct, you got it wrong by saying A + C = B + D while in fact the right thing to say would be A^2 + C^2 = B^2 + D^2 which is just the british flag theorem if you select the radius of the circle to be 0, since the radius can be any number, 0 is also a valid choice.
Btw, sqrt(7361) is about 85.796. It makes sense that the difference between the two long ones is less than the difference of the two short ones.
Awesome
ahm....from the drawn you are assuming that u can make a 90 degree angle line to the center of the circle...but that isn't "evident" on the draw, but i guess it's not possible to solve if that is not the case.
The problem says they're *tangent* lines to the circle. That means each one is, er, flat against a point on the circle and at right angles to the radius at that point.
Great
Gran problema de olimpiadas excelente like
who's hungry
Me
哈哈 听印度朋友讲数学
87
Hay thật
i like
CLAP CLAP CLAP !
Seramalatuhgn enkuwan yanten x lifelglh qerto righthnm alfelg deffar azza eski leqeqegnna braseh zabbiya bicha nurrr ynenen x ante aydelehm mitnegregn kante shi etf ymibeltu ngrewgnalna begeza ejh atqlel bayhon yanten x let me tell you yante x yaahya sega alga silut amed ytebalew is best fit for you ok hule khusr ywurre tilq yelewm ytebalew haq new tilq millionair bileh litakebrew sitl bymnderu ltelkasha were endahyaw sega silkesekes setagegnew yane new ekkhkhkjjhkhkhkhjgeeeey miyasegnhh ene badaf endashagn tefeqdolnal ekhkjkjkhkhk allah zerzuryahn ezaaae allah yargewna ameen😮😮😮
眼看了，但？？？？？？？？？？？？／～～！！＠＃＃＄＄％％
Just a little too tedious, along with the not-so-easy-on-the-ear English pronunciation - the video could be almost half its length without losing everything, and not every note written must also be narrated.
The solution is nice and beautiful - but I think you could also give 10-20 seconds before solving, on the way you looked at the problem, and how you attacked it, and show the viewers your "war plan", not just the technical details narrated so indifferently - that one KNOWS you had the solution complete to start with. Give it some life! let children not only see how you solve, but develop an appetite for STRUGGLING with such questions.
Thank you anyway.
Why? I don't care!
run the following in bbcbasic and change the number for "r=" in line 20:
10 print "brain station end result will shock you":dim x(1,3),y(1,3)
20 l1=33:l3=85:l4=35:r=25:la=60:sw=la/47:sw=2*pi/59:print la:rem @zoom%=1.4*@zoom%
30 s1=sqr(l1^2+r^2):s3=sqr(l3^2+r^2):s4=sqr(l4^2+r^2):print s1,s3,s4:goto 50
40 lhu=(d1^2-d2^2+d3^2)/2/d3:hu=sqr(d1^2-lhu^2):return
50 d1=s1:d2=s4:d3=la:gosub 40:xm=hu:print "xm=";xm:disy=r*r+l1^2-xm^2:if disy0 then w1=w else w2=w
180 if abs(dw)>1E-10 then 170 else return
230 w=sw:xu=0:yu=0:gosub 110:x(1,0)=xs:y(1,0)=ys:w=w+sw:xu=lb:gosub 110:x(1,1)=xs:y(1,1)=ys
240 w=w+sw:yu=la:gosub 110:x(1,2)=xs:y(1,2)=ys:w=w+sw:xu=0:gosub 110:x(1,3)=xs:y(1,3)=ys
250 x(0,0)=0:y(0,0)=0:x(0,1)=lb:y(0,1)=0:x(0,2)=lb:y(0,2)=la:x(0,3)=0:y(0,3)=la
260 xmin=0:xmax=lb:ymin=xm-r:ymax=xm+r
270 masx=1200/lb:masy=850/la:if masx
87
87