Math Olympiad | Can you find the length AB? | (Justify your answer) |

[@Vakyanshpro]

You explains very nicely

[@PreMath]

Thanks a lot 😊🌹

[@harrymatabal8448]

Excellent. Your videos make you think ❤

[@josefpichler7271]

Nice, with your help I get back a lot of knowledge learned long times ago ...

[@jamestalbott4499]

Thank you!

[@alster724]

Impressive!

[@wackojacko3962]

@ 1:47 there is a phenomenon of the cathode ray tube of old televisions is known as "blacker than black" . I never thought of what appears to be the illusion of "whiter than white" of the color change. And of course a shout out to using the Intersecting Chord Theorem. 🙂

[@PreMath]

😀 Thanks for the feedback ❤️

[@Shell1950]

I solved it days before my 75th birthday. Never stop learning.

[@Mediterranean81]

AP=a ; EQ=b intersecting chords theorem b(a+7)=5(12) a+7=60/b a=-7+ 60/b (#) a=(-7b+60)/b intersecting chords theorem again 6*10=a(b+7) 60=(60-7b)(b+7)/b 60b=60b+420-7b^2-49b 0=-7b^2-49b+420 0=b^2+7b-60 you can use quadratic formula but i'm factoring 0=b^2+12b-5b-60 0=b(b+12)-5(b+12) 0=(b-5)(b+12) b-5=0 or b+12=0 b=5 or b=-12 b>0 so b=5 from equation (#) a=-7+(60/b) a=-7+(60÷5) a=-7+12 a=5 a=5 ; b=5 so AE=a+b+7=5+5+7=17

[@PreMath]

Excellent! Thanks for sharing ❤️

[@yalchingedikgedik8007]

Thanks Sir Very well and enjoyable With my respects ❤❤❤

[@quigonkenny]

Bu intersecting chords theorem, for any two intersecting chords in a circle, the products of the lengths either side of the point of intersection is the same. In this case, CP•PD = AP•PB and EQ•QF = AQ•QB. Let AP = x and QB = y. CP•PD = AP•PB 6(10) = x(7+y) 60 = 7x + xy xy = 60 - 7x y = (60-7x)/x EQ•QF = AQ•QB 5(12) = (x+7)y 60 = (x+7)y 60 = (x+7)(60-7x)/x 60x = 60x + 420 - 7x² - 49x 7x² + 49x - 420 = 0 x² + 7x - 60 = 0 (x+12)(x-5) = 0 x = 12 ❌ | x = 5 y = (60-7(5))/(5) y = 25/5 = 5 AB = x + 7 + y = 5 + 7 + 5 = 17

[@PreMath]

Excellent! Thanks for sharing ❤️

[@batavuskoga]

Using the chords theorem, it's easy to find the solution

[@PreMath]

Thanks for the feedback ❤️

[@zdrastvutye]

this revealed to need 2 nested calculations, but in the end i have the same result, despite i did not apply any theorem, only by calculating distances: 10 print "premath-can you find the length ab":dim x(9),y(9):@zoom%=1.4*@zoom% 20 l1=6:l2=10:l3=7:l4=5:l5=12:sl=l1+l2+l3+l4+l5:sw=l1^2/sl:r=sw:goto 170 30 xm=-sqr(abs(r*r-(ye-ym)^2)):goto 140 40 yp=-sqr(l3^2-xp^2):goto 80 50 dyc=2*yc*(ym-yp):zx=r*r-l1^2-xm^2-ym^2+xp^2+yp^2+dyc:if xp=xm then stop 60 xc=zx/2/(xp-xm):dgu1=(xc-xp)^2/sl^2:dgu2=(yc-yp)^2/sl^2 70 dgu3=l1^2/sl^2:dg=dgu1+dgu2-dgu3:return 80 yc=-sl:gosub 50 90 dg1=dg:yc1=yc:yc=yc+sw:yc2=yc:if yc>sl then return 100 gosub 50:if dg1*dg>0 then 90 110 yc=(yc1+yc2)/2:gosub 50:if dg1*dg>0 then yc1=yc else yc2=yc 120 if abs(dg)>1E-10 then 110 else return 130 goto 140 gosub 40:dxd=(xp-xc)*(l1+l2)/l1:xd=dxd+xc:dyd=(yp-yc)*(l1+l2)/l1 150 yd=dyd+yc:dfu1=(xd-xm)^2/sl^2:dfu2=(yd-ym)^2/sl^2:dfu3=r*r/sl^2 160 df=dfu1+dfu2-dfu3:return 170 ym=(l4-l5)/2:ye=l4:xp=-l3: gosub 30 180 df1=df:xp1=xp:xp=xp+sw:if xp>l3 then else 200 190 r=r+sw:goto 170 200 xp2=xp:gosub 30:if df1*df>0 then 180 210 xp=(xp1+xp2)/2:gosub 30:if df1*df>0 then xp1=xp else xp2=xp 220 if abs(df)>1E-10 then 210:rem die schnittpunkte berechnen 230 xq=0:yq=0:dx=xp-xq:dy=yp-yq:x1=xq:y1=yq:dl=l3 240 kxx1=(dx/dl)^2:kxx2=(dy/dl)^2:kxx=kxx1+kxx2:kx1=(x1-xm)*dx/dl:kx2=(y1-ym)*dy/dl 250 kn=(x1-xm)^2+(y1-ym)^2-r*r:kx=kx1+kx2:kx=kx/kxx:kn=kn/kxx 260 dis=kx^2-kn:if dis

[@PrithwirajSen-nj6qq]

Intercepting chords theorem and arithmetic ------_--------- Sum of the segments of AB is fixed as AB is fixed. The larger segment is greater than 7 according to the figure. The product of segments is 60. ++ Hence we have to break 60 into two parts so that the product is 60.And the difference between larger and smaller will be7 Hence 60=10*6(case 1) =12*5 ( case2) = 15*4 (.case 3) Case 1 If larger segment is 10 ,then the smaller segment will be (10-7)=3 10*3 is not equal to 60 Hence this break up may not be accepted. Case 3 if the larger segment is 15, then the smaller section will 15-7=8 Product of 15*8 exceeds 60 Hence this break up may not be accepted. Case 2 If the larger segment is 12 and the smaller will be 12-7=5. The product of 12 and 5 is 60 This break up must be accepted . Comment please. Length of AB = 12+5=17

[@gelbkehlchen]

Solution: According to the tendon theorem: (1) AP*(7+QB) = 6*10 ⟹ (2) (AP+7)*QB = 5*12 ⟹ (1a) 7*AP+AP*QB = 60 ⟹ (2a) AP*QB+7*QB = 60 ⟹ (1a)-(2a) = (3) 7*AP-7*QB = 0 ⟹ (3a) AP = QB in (1a) ⟹ (1b) AP²+7*AP = 60 |-60 ⟹ (1c) AP²+7*AP-60 = 0 |p-q formula ⟹ (1d) AP1/2 = -3.5±√(3.5²+60) = -3.5±8.5 ⟹ (1e) AP1 = -3.5+8.5 = 5 and (1f) AP2 = -3.5-8.5 = -12 [invalid in geometry] ⟹ AB = 2*AP+7 = 2*5+7 = 17

[@Tmwyl]

I got this one!

[@marcgriselhubert3915]

I dd it just like you. Another possible method?? (I am always surprised by the way you solve second degree equations)

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@sorourhashemi3249]

Thanks. AP=a and QB=b, (7+a)b=60== >b=60/7+a, and (7+b)a=60, ===>a=60/7b, so (7+60/7+b)b=60===>7b^2+49b-420==> b=5 and a=5 as well and AB=17.

[@AmirgabYT2185]

AB=17

[@PreMath]

Excellent! Thanks for sharing ❤️

[@santiagoarosam430]

Potencia de Q =5*12=60=(7+a)b Potencia de P =6*10=60=(7+b)a 7b+ab=7a+ab ; a=b ; 60=7a+a^2 ; a=5. AB=7+2a=17. Gracias y saludos.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@phungpham1725]

La bel the two unknown red line segment as x and y. By using chord theorem, we can find x= y So, x(7+x) =60--> sqx +7x-60=0 --> x= (-7+17)/2 =5 ( negative result rejected) AB= 17😅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@kalavenkataraman4445]

AB = 17 units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@himadrikhanra7463]

Ab= x 60= x × ( x - 7) 60 = x ×( x - 7) AB =12 ?

[@prossvay8744]

Let AP=x ; BQ=y x(y+7)=(6)(10) xy+7x=60 (1) y(x+7)=(5)(12) xy+7y=60 (2) (1) and (2) xy+7x=xy+7y 7x=7y So x=y (1): x^2+7x-60=0 so x=5 ;y=5 so AB=5+7+5=17units.❤❤❤.Best regards.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@LuisdeBritoCamacho]

Assuming that Length of Line AB = (X + 7 + Y) Linear Units. Making use of the Intersecting Chords Theorem: "The Intersecting Chords Theorem states that when two chords of a circle intersect within the circle, the product of the segments of one chord is equal to the product of the segments of the other chord." 1) Given the System of Two Nonlinear (Curvilinear) Equations with Two Unknows (X ; Y): 2a) (6 * 10) = X * (7 + Y) : 60 = X * (7 + Y) 2b) (5 * 12) = Y * (7 + X) ; 60 = Y * (7 + X) 3) Solutions (Intercepting Point): X = 5 and Y = 5 4) (X + 7 + Y) = 5 + 7 + 5 = 17 lin un 5) ANSWER : Length of Line AB is equal to 17 Linear Units.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@unknownidentity2846]

Let's find the length of AB: . .. ... .... ..... It seems that we have to apply the intersecting chords theorem: CP*DP = AP*BP EQ*FQ = AQ*BQ 6*10 = AP*BP 5*12 = AQ*BQ 60 = AP*BP 60 = AQ*BQ ⇒ AP*BP = AQ*BQ AP*BP = AQ*BQ AP*(BQ + PQ) = (AP + PQ)*BQ AP*BQ + AP*PQ = AP*BQ + PQ*BQ AP*PQ = PQ*BQ ⇒ AP = BQ 60 = AP*BP = AP*(BQ + PQ) = AP*(AP + PQ) = AP*(AP + 7) = AP² + 7*AP AP² + 7*AP − 60 = 0 AP = −7/2 ± √[(7/2)² + 60] = −7/2 ± √(49/4 + 60) = −7/2 ± √(49/4 + 240/4) = −7/2 ± √(289/4) = −7/2 ± 17/2 Since AP>0, the only useful solution is: AP = −7/2 + 17/2 = 10/2 = 5 AB = AP + PQ + BQ = 2*AP + PQ = 2*5 + 7 = 17 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Mediterranean81]

AP=a ; EQ=b intersecting chords theorem b(a+7)=5(12) a+7=60/b a=-7+ 60/b (#) a=(-7b+60)/b intersecting chords theorem again 6*10=a(b+7) 60=(60-7b)(b+7)/b 60b=60b+420-7b^2-49b 0=-7b^2-49b+420 0=b^2+7b-60 you can use quadratic formula but i'm factoring 0=b^2+12b-5b-60 0=b(b+12)-5(b+12) 0=(b-5)(b+12) b-5=0 or b+12=0 b=5 or b=-12 b>0 so b=5 from equation (#) a=-7+(60/b) a=-7+(60÷5) a=-7+12 a=5 a=5 ; b=5 so AE=a+b+7=5+5+7=17

[@PreMath]

Excellent! Thanks for sharing ❤️