Russian Math Olympiad | A Very Nice Geometry Problem

Russian Math Olympiad | A Very Nice Geometry Problem | Square inside a semicircle
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[@contnuum1607]

ABC is right angled triangle with angle C = 90 deg. OE = sqrt(5) => OD = sqrt(10) = radius => AB = diameter = 2 X OD = 2sqrt(10). Since ABC is right angled triangle with angle(ACB) = 90 deg => BC = AB X cos(angle(ABC)) = 2 X sqrt(10) X sqrt(10/15) = 2 X sqrt(4 X 5/3) = 4/3 X sqrt(15)

[@hoehlengnarf7496]

I think you left out one step: angle(ABC) = angle(OBE) and angle at O (in OBE) is 90°, hence cos(angle(ABC)) = cos(angle(OBE)) = sqrt(10) / sqrt(15) = sqrt(10/15). I would also consider this easier than the argument in the video as I must admit I don't know (or at least I don't remember) the theorem applied to the lines involving M and N.

[@venkateswararaokotaru4302]

Good 👍 .

[@venkateswararaokotaru4302]

After getting OE=√5 , RADIUS OD=OA= OB=√10 and BE=√15 , We Have ∆ABC ~ ∆EBO Give us BC/AB=OB/BE So BC = (2√10)×√10/√15 = 4/3 √15 .

[@hongningsuen1348]

One method to solve the problem is to use proportionality equation for corresponding sides of similar triangles ACB (constructed) and EBO after finding radius of the semicircle. Notes for students: Whenever you see a semicircle, equal radii and Thales theorem should come to your mind. Whenever you see right-angled triangles, Pythagoras theorem and similarity of triangles hence proportionality equation of corresponding sides should come to your mind.

[@dmitrymelnik8296]

Let ABC = \alpha. Apparently, the triangle OCB is isosceles with two sides equal to R, and the angle between them is (180 - 2*alpha). We can either invoke the cosine theorem, or recognize that BC = 2* R*cos(\alpha). From EOB we find cos(\alpha) = \sqrt(2/3). The rest is arithmetic.

[@jpl569]

Here is another proof… Let H be the projection of C on AB, and ß the angle EBO (also CBH). Obviously, the radius of the circle is R = √10. Then tg ß = √5 / √10 = 1 / √2, and sin ß = CH/X, and cos ß = R/BE. And also : tg ß = CH / BH. As angle COH is 2ß (nice property of the circle !), then : X = CH / sin ß = R sin 2 ß / sin ß = 2 R cos ß = 2 R^2 / BE. Then X = 2 . 10 / √15 = 4 √15 / 3. Thank you for your videos !! 🙂

[@varathan3558]

at 7:40 it simpliest to draw the CA line and take cos(OBE)=cos(ABC)! The solution just appears in front of you!

[@RAG981]

For the ending I used similar triangles ABC and EBO rather than chord properties. It was simpler to do.

I did this same way 😁

[@HCHLEE2007]

Yes much simpler

[@kateknowles8055]

(r - sqrt(5))(r+sqrt(5)) = r.r -5 with OE produced to the circumference at G so that OG = radius =r. Now if EO is produced also to meet a reflection (H) of G in diameter AB, the product HE.EG is the r.r-5 mentioned at the start. HE.EG = ED.ED by the intersection of HG and the chord DE extended. so r.r -5=5 so r.r =10 so r=sqrt(10) Now to notice that CE.EB also = ED.ED by intersection of chords so this is 5 also From triangle OBE, Pythagoras' theorem gives : EB.EB = OB.OB+ OE.OE = r.r + 5 = 15 CE = 5/sqrt(15) EB = sqrt(15) so BC = 5/sqrt(15) + sqrt(15) = 5/15 times (sqrt(15) ) +(15/15) sqrt ( 15 ) = 4/3 of sqrt (15)

[@MorgKev]

Also by intersecting chords: DE.DE = CE.EB... this is marginally quicker and a bit less messy.

[@jonpress6773]

Use coordinate geometry, placing the O at the origin. The equation of the circle is x^2 + y^2 = 10. It's easy to determine the coordinates of points B and E from the givens, so use them to find the equation of line BCm which turns out to be y=x/sqrt(2) + sqrt(5). Solve the two equations together to get the coordinates of point C, which turns out to be (-sqrt(10)/3, 4/3 sqrt(5)). Now we have the coordinates of B and C, so the distance formula gives the answer. Not as elegant as Math Booster's solution, but it still works.

[@BartvandenDonk]

It is far more elegant than his solution. 😉

[@MrPaulc222]

r = sqrt(10) as it's the square's diagonal. sqrt(5)^2 + sqrt(10)^2 = (EB^2, so EB = sqrt(15) BCA and BEO are congruent. (EB)/(OB) = (AB)/(CB) Call (CB), x as it's the target value. (sqrt(15))/(sqrt(10)) = (2*sqrt(10)/x Cross multiply: (sqrt(10)) * (2*sqrt(10)) = (sqrt(15)*x 20 = (sqrt(15)*x 400 = 15x^2 Reduce: 80 = 3x^2 (80/3) = x^2 (sqrt(80))/(sqrt(3) = x. Rationalise to (sqrt(240))/3 = x Simplify: (4*sqrt(15))/3 = x I have 5.164(rounded to 3dp)

[@hongningsuen1348]

Method 1 using Thales theorem and similar triangles: 1. Let BC be x. 2. Side of square OEDF = √5 (property of square) Hence diagonal of square OD = √[(√5)^2 + (√5)^2)] = √10 (Pythagoras theorem) Hence AO = BO = OD = √10 (radii of semicircle) 3. Draw AC to form ∆ABC. Angle ACD = 90 (Thales theorem) 4. In ∆ACD AC^2 = AB^2 - BC^2 (Pythagoras theorem) = (AO + BO)^2 - x^2 = (√10 + √10)^2 - x^2 = 40 - x^2 AC = √(40 - x^2) 5. ∆ABC ~ ∆EBO (AAA) Hence AC/EO = BC/BO [√(40 - x^2)]/√5 = x/√10 (40 - x^2) = x^2/2 3x^2 = 80 x^2 = 80/3 x = √80/√3 = (√3√80)/3 = √3 √(16 x 5)/3 = (4√15)/3 Method 2 using intersecting chord theorem: 1. Side of square OEDF = √5 (property of square) Hence diagonal of square OD = √[(√5)^2 + (√5)^2)] = √10 (Pythagoras theorem) Hence AO = BO = OD = √10 (radii of semicircle) 2. In ∆BOE BE^2 = BO^2 + OE^2 = 10 + 5 BE = √15 3. Extend DE to G on arc BC to form chord DG. DE = GE = √5 (OE is perpendicular bisector of chord DG from centre.) 4. For chords BC and DG intersecting at E DE x GE = BE x EC (intersecting chord theorem) √5 x √5 = √15 x EC Hence EC = √(5/3) 5. BC = BE + EC = √15 + √(5/3) = (4√15)/3

[@lusalalusala2966]

Draw a perpendicular line to CB from O, which intersects it at the midpoint T since CB is a chord. Next the area of the triangle OEB is sqrt(10)*sqrt(5)/2 =EB*OT/2. Therefor OT= sqrt(10)*sqrt(5)/sqrt(15)=sqrt(10/3). Finally CB=2 TB, where TB=sqrt(OB^2-OT^2)=sqrt(10-10/3)=sqrt(20/3)=2sqrt(5/3), ie., CB=4sqrt(5/3).

[@marcgriselhubert3915]

The side length of the square is sqrt(5) and the radius of the circle is sqrt(2).sqrt(5) = sqrt(10) We us an orthonormal, center O, first axis (OB). The equation of the circle is x^2 + y^2 = 10 We have B(sqrt(10);0) and E(0; sqrt(5)), then VectorBC(-sqrt(10); sqrt(5)) is colinear to VectorU(-sqrt(2); 1) The equation of (BE) is: (x -sqrt(10)).(1) - (y).(-sqrt(2)) = 0 or x + sqrt(2).y -sqrt(10) = 0, or x= -sqrt(2).y +sqrt(10) C is the intersection of (BE) and the circle, the ordinate of C is such as: (-sqrt(2).y +sqrt(10))^2 + y^2 = 10, or 3.y^2 -4.sqrt(5).y = 0 So the ordinate of C is (4.sqrt(5))/3, and its abscissa is -sqrt(2). (4.sqrt(5))/3) + sqrt(10) = (-4.sqrt(10))/3 + sqrt(10) = -sqrt(10)/3 Finally we have C(-sqrt(10)/3; (4.sqrt(5))/3) and Vector BC(-4.sqrt(10))/3; (4.sqrt(5))/3) and BC^2 = 160/9 + 80/9 = 240/9 Finally BC = sqrt(240)/3 = (4.sqrt(15))/3.

[@MarieAnne.]

My solution started out the same Find side length of square = √5 Find radius of circle = √10 Find length of BE = √15 But instead of completing the circle, I joined A to B to form △ABC We find that △ABC ~ △EBO by AA ∠ACB = 90° (angle subtended by diameter = 90°) and ∠EOB = 90° ∠ABC = ∠EBO (same angle) Using similar triangles we get: BC/BO = AB/EB BC/√10 = 2√10/√15 *BC = 20/√15 = 4√15/3*

[@jandirpassos5327]

I bounced the square to the right side and applied the same properties as the final part of the solution used in the video.

[@zdrastvutye]

i have calculated repeatedly the deviation of point c from the circle with interpolation: 10 print "mathbooster-russian math olympiad":a1=5:l1=sqr(a1) 20 dim x(3,2),y(3,2):r=l1*sqr(2):xb=2*r:yb=0:yd=l1:@zoom%=1.4*@zoom% 30 xd=r-sqr(r*r-yd^2):xe=xd+l1:ye=yd:xb=2*r:yb=0:ye=l1:sw=.1:goto 70 40 dxk=(xe-xb)*k:xc=xb+dxk:dyk=(ye-yb)*k:yc=yb+dyk 50 dgu1=(xc-r)^2/a1:dgu2=yc^2/a1:dgu3=r*r/a1:dg=dgu1+dgu2-dgu3 60 return 70 k=sw:gosub 40 80 dg1=dg:k1=k:k=k+sw:k2=k:gosub 40:if dg1*dg>0 then 80 90 k=(k1+k2)/2:gosub 40:if dg1*dg>0 then k1=k else k2=k 100 if abs(dg)>1E-10 then 90 110 lg=sqr((xc-xb)^2+(yc-yb)^2):print "der abstand BC="; lg 120 x(0,0)=xd:y(0,0)=0:x(0,1)=x(0,0)+l1:y(0,1)=0:x(0,2)=xd:y(0,2)=l1 130 x(1,0)=x(0,0)+l1:y(1,0)=0:x(1,1)=x(1,0):y(1,1)=l1:x(1,2)=xd:y(1,2)=yd 140 x(2,0)=xe:y(2,0)=ye:x(2,1)=xc:y(2,1)=yc:x(2,2)=xd:y(2,2)=yd 150 x(3,0)=r:y(3,0)=0:x(3,1)=2*r:y(3,1)=0:x(3,2)=xe:y(3,2)=ye 160 masx=1000/2*r:masy=700/r:if masx run in bbc basic sdl and hit ctrl tab to copy from the results window

[@tgx3529]

This problem can also be solved by older students with the help of Thalet's circle - the angle at C is right. The Pythagorean theorem gives 40= k^2+(z+sqrt15)^2 where k is the line connecting point C to the diameter. sinα is in the small triangle sqrt5/sqrt15=sqrt3/3 sin's theorem for a right triangle above the diameter gives sinα/k=sin90/2sqrt10... it leads to the same result of sqrt(80/3)-sqrt15 for the shorter segment of the unknown segment.

[@LuisdeBritoCamacho]

As in all Geometrical Problems, the Real Problem is: Where to look? In this particular case one must look to Line OD and understand that the Diagonal of the Square is equal to the Radius of the Semicircle. So, if the side of the Square is equal to sqrt(5); (sqrt(5) * sqrt(5) = 5; then its Diagonal is equal D^2 = sqrt5)^2 + sqrt(5)^2. D^2 = 5 + 5 = 10. So, Diagonal is equal to sqrt(10) ~ 3,2 Linear Units. Note that the Diagonal of any Square is always equal to : D = Side*sqrt(2). In this case sqrt(5) * sqrt(2) = sqrt(10); as staed before!! Radius = sqrt(10) and, EB^2 = 5 + 10 ; EB^2 = 15 ; EB = sqrt(15) ~ 3,9 lin un By the Theorem of Similarity between Triangles we have : BO/EB = BC'/BC : C' is the Middle Point between OF sqrt(10) / sqrt(15) = (sqrt(10) + (sqrt(5)/2) / BC 3,16/3,87 = (3,16 + 1,12) / BC 0,82 = 4,28 / BC BC = 4,28/0,82 BC = 5,22 Linear Units, approximately.

[@mickodillon1480]

Exactly. I missed that somehow! Therefore I found it really hard.

[@LuisdeBritoCamacho]

@@mickodillon1480 , don't worry. Be happy!

[@quigonkenny]

Let r be the radius of the circle and s be the side length of the square. As square OEDF has area 5, its side length is the square root of that area, or √5. Draw radius OD. In addition to being a radius of the semicircle, OD is also a diagonal of the square. As such, it's length is √2 times the side length. r = √2•√5 = √10. Triangle ∆EOB: EO² + OB² = BE² (√5)² + (√10)² = BE² BE² = 5 + 10 = 15 BE = √15 Draw CA. As C is on the circumference of the semicircle and is the angle between the ends of the diameter AB, ∠ C = 90°. As ∆EOB and ∆BCA share angle ∠B and are both right triangles, ∆EOB and ∆BCA are similar. Triangle ∆BCA: BC/AB = OB/BE BC/2√10 = √10/√15 = √2/√3 BC = (2√10)(√2/√3) BC = 4√5/√3 = 4√15/3 ≈ 5.164

[@maxgladkikh2326]

Я помню эту задачу! 9 класс советской школы, 40 лет назад. Почти все справились без затруднений. Как она может быть олимпиадной?

[@yadonghu6219]

连接A,C;三角形ACB为直角三角形。直角三角形ACB与EOB相似，对应边成比例。BC：sqrt(10)=2*sqrt(10)：sqrt(15)。BC=20/sqrt(15)。

[@marioalb9726]

A = ½R² = 5 cm² R = √10 cm tan α = s/R = √5 / √10 = 1/√2 α = 35,26° x = 2R cos α x = 5,164 cm ( Solved √ )

[@민동익-y3j]

닮음을 이용하면 선분 BC의 중점을 M이라 했을 때 선분 EB:선분OB=선분OB:선분BM이므로 선분 BM의 길이는 sqrt(20/3) 선분 BC의 길이는 4sqrt(5/3)임을 알 수 있습니다

[@tituschirila7650]

after finding BE you can find height from O to BE = OP (equaling area of triangle OBE) and from that half of CB (OP perpendicular from center divides OB by two) and from that by pythagora PB. PB is hlf BE - simple as that.

[@ST-sd8un]

Excelente! OP (raíz 15)= (raíz 5)(raíz 10) OP= [(raíz 5)(raíz 10)]/[(raíz 5)(raíz 3)] OP= raíz 10/raíz 3 OP=(raíz 30)/3 Pitagoras BP^2 + [(raíz 30)/3]^2 = (raíz 10)^2 BP^2 = 10 - (30/9) BP^2 = 60/9 BP= 2(raíz 15)/3 BC = 2 BP BC = 4 (raíz 15)/3

[@timc5768]

Perhaps also: By similar triangles DFA and BFD ( equivalent to using intersecting chords theorem, I think): [r +sqrt(5))/sqrt(5) = (sqrt(5))/[r - sqrt(5)], where 'r = radius', so ' r^2 - 5 = 5', and 'r = sqrt(10)', so BE = sqrt(15) , (by Pythag), and 'AB = 2sqrt(10)' Then by similar triangles EOB and ACB : [OB/BE) = [CB/BA] , (= cos(B)), so CB = [2sqrt(10)][{sqrt(10)}/sqrt(15)}] = 2sqrt(20/3) = 4sqrt(5/3) = (4/3)sqrt(15)

[@TheAndreArtus]

That is the same way I went with it.

[@venkateswararaokotaru4302]

After getting OE=√5 , RADIUS OD=OA= OB=√10 and BE=√15 , We Have ∆ABC ~ ∆EBO Give us BC/AB=OB/BE So BC = (2√10)×√10/√15 =(4/3) √15 .

[@JamesCalabut]

Triangle BOE is similar to BCA, so sqrt(15)/sqrt(10) = 2*sqrt(10)/x. Thus x=2*10/sqrt(15). Simplifying: x=2*10*sqrt(15)/15=4/3*sqrt(15)

[@murdock5537]

φ = 30°; area ∎DFOE = 5 → FO = a = √5 → DO = a√2 = r = √10 ∆ ABC →sin(BCA) = 1 → AB = 2r ∆ BEO → EO = r/√2; BO = r → BE = r√6/2 → ABC = δ → cos⁡(δ) = 2r/r√6 = √6/3 = BC/2r → 3BC = 2r√6 → BC = (2/3)r√6 = 4√15/3

[@ExpressStaveNotation]

Extend DE to make the chord that intersects with chord BC. R5.r5 = BE.EC 5 = r15.EC EC = 5/r15 BC = 5/r15 + r15.

[@magdyeldajany9973]

Extend the chord DE to intersect the circle at point L. From the two chords CB and DL, we can find CE

[@grzegorzmordas9134]

The radius of the circle is Sqrt(2x5) = Sqrt(10). Triangles BEO and ABC are similar. Thus: OB/BE=AB/BC or r/Sqrt(r^2+5) = x/2r. Substituting r=Sqrt(10) we get: Sqrt(10)/Sqrt(15) = x/2Sqrt(10), hence: x=20/Sqrt(15) = 4Sqrt(15)/3.

[@bakrantz]

Triangle ABC is a right triangle and similar to triangle BOE. So X can be solved for by similarity.

[@dlspark7965]

BC = (4/3)*sqrt(3A) where A = Area of square OEDF

[@constantinfedorov2307]

Я не очень понимаю, как такая элементарная задачка могла попасть на олимпиаду. В условии дано все - радиус известен, это диагональ квадрата, отрезок BE тоже считается тривиально (стороны EO = √5 OB = √10 => BE = √15), и дальше опять тривиальное подобие треугольников ABC и EBO. BC/AB = OB/EB; BC = (2√10)√(10/15) = (4/3)√15; Тут олимпиадой и не пахнет, на ЕГЭ бывают задачи сложнее.

[@KipIngram]

When I mention x and y coordinates, it is with the center of the circle taken as origin. The square has side length sqrt(5), so the circle has radius sqrt(2)*sqrt(5)=sqrt(10). So the angle X makes with the horizontal axis at B is arctan(sqrt(5)/sqrt(10)) = arctan(1/sqrt(2)). We can now use the law of sines to get the vertical coordinate of C: 2*y(C)/sin(2*arctan(1/sqrt(2)) = 2*sqrt(10) y(C) = sqrt(10)*sin(2*arctan(1/sqrt(2)) y(C) = 2.9814 Now we can use these facts to get the x coordinate of C: 2.9814/(sqrt(10)-x) = 1/sqrt(2) sqrt(10)-x = 2.9814*sqrt(2) x = sqrt(10) - 2.9814*sqrt(2) x = -1.0541 Finally we can use the Pythagorean theorem to get X: X^2 = (sqrt(10)+1.0541)^2 + (2.9814)^2 X^2 = 26.666 X = 5.1640 Q.E.D.

[@santiagoarosam430]

a²=5→ a=√5→ r= Diagonal del cuadrado =a√2=√5√2=√10→ EB²=(√5)²+(√10)²=15→ EB=√15→ Si G es la proyección ortogonal de O sobre EB y h=OG→ h√15=√5√10→ h=√50/√15 → Razón de semejanza entre los triángulos ACB y OGB = s=AB/OB=2r/r=2→ AC=2h→ CB²=AB²-AC²=(2r)²-(2h)²=(2√10)²-(2√50/√15)²→ CB=20√15/15 =4√15/3. Gracias y saludos.

[@vcvartak7111]

You can also join AC and ACB and BOE are similar triangles and take side proportion

[@arulbiswas1260]

Such ingenious thinking!!!!

[@hongningsuen1348]

Good use of Thales theorem.

[@NorArt-mb3lv]

I did it the same way.

[@gaylespencer6188]

Found angle CBA. Then took Cos (CBA) and multiplied it by the diameter of (2*10^.5) = 5.16......

[@plamenpenchev262]

Put the same square to the right. Chord property then Sqrt(15)×|CE| = sqrt(5)×sqrt(5) Then sqrt(15) + sqrt(15)/3

[@alexeygourevich6967]

Радиус окр-ти DO = √2 * DE = √2 * √5 = √10 = ОВ. Пусть угол ЕВО равен β. Тогда tgβ =√5/ √10 = 1/√2. Но треуг АСВ - прямоугольный с углом С=π/2, поэтому АС/СВ = tgβ = 1/√2. Также AC^2+ CВ^2 = (2√10)^2 = 40, отсюда, поделив на CВ^2, получим (1/√2)^2 + 1 = 40/ CВ^2, или 3/2 = 40/ CВ^2, отсюда CВ^2 = 80/3, т.е. CВ = √(80/3) = √(16*5/3) = 4* √((3*5)/(3*3)) = 4*√15 / 3.

[@georiashang1120]

R=BO=DO=CO=FE=√10 ; EO=√5 ; cos(∠EBO)=√2/√3 ; BC*(1/2)=2√5/√3 ; BC=4√5/√3=4√15/3

[@mg.1044]

metot 2 ACB ~ OEB sin ß = CB /2r = r/ EB

[@JPTaquari]

I did it by triangulation and then by similarity of triangles and arrived at 5.24228. I must have rounded up a little. Nice exercise, congratulations From Brasil !

[@UrievJackal]

We don't really know, which level of math, this Olympiad demands. Therefore, when first time I used a sine for BOE triangle, maybe I was wrong. Maybe children of that Olympiad may not know of trigonometry. So the best solution is to use similar triangles BOE and BCA at the end. Thus, BO/BE = x/AB; R/BE = x/2R.

[@retarody9542]

Cos of tria = √10 / √15 2r = 2√10 X = cos x 2r = 4√15 / 3

[@user-gn4mq5cs6e]

Drop a perpendicular on X, it will cut at x/2 and calculate X straight away

[@sakibalmahmud686]

Triangle ABC and Triangle OBE symmetric.. Then OBE is fully known..then ABC aslo known..from that..

[@احمدالجوهر-ظ4ت]

OB×AB=CB×EB ACB right angled

[@1ciricola]

The radius of the circle is the length of line segment DO, which is √10 , since the length of each side of the square is √5. A line drawn between C and O likewise has a length of √10 ΔCOB is an isosceles triangle. Since line segments EO and OB are known [√5 & √10 respectively], arctan .707 = 35.26° The isosceles triangle has two angles of 35.26° ∠CBO and ∠BCO The Altitude bisects the base BC at M. The length of each half can be calculated using the cosine function. Cos 35.26° = BM/BO = BM/√10 (.816)(√10) = BM = 2.58, length of BC = (2)(BM) = 5.16

[@InventPeace1]

Did they leave out of the description that " O " was the bisection point of AB ? InventPeaceNotWar

[@joyhuang7845]

有另一左右對稱正方形，畫出來之後，也是利用圓內幕性質，即可求出EC

[@shoebmd428]

Triangle ACB and OEB are similar, , it could have saved some steps

[@اقرء]

Tanks for watching

[@Cricketdoctor_1999]

🤙Nice geometry problem bro!...

[@tamirerez2547]

Very nice solution 👍

[@himadrikhanra7463]

14+ 4root 5....? Line construction...similarly...diameter...Pythagoras

[@wiwa5613]

|FD| = |OF| = sqrt(5) |OD| = r = sqrt(5) x sqrt(2) = sqrt (10) |BE| =sqrt (sqrt(10))^2 + (sqrt(5))^2) = sqrt(15) P(triangle) OBE = 1/2 x |OB| x |OE| = 5/2sqrt(2) and P(triangle) OBE = 1/2 x h(trangle BOE) x |BE| = 5/2sqrt(2) => h(trangle BOE) = 1/3 x sqrt(30) |OB| = |OC| = r = sqrt(10) => r^2 = (h (trangle BOE))^2 +(1/2 x |BC| )^2 => |BC| = 4/3 x sqrt(15)

[@michallesz2]

IF BD=R then AO=OB and OD=R ERROR

[@vrcfncpdci]

A good problemi for mr. perrelman the russian genius

[@Dauooria]

매일 매일 재미있는 수학 영상을 올려주셔서 감사합니다! I'm korean student

[@just_isaac15]

To the people who are stating other methods, I am so jealous😭😭

[@WlodekCiejkaTV]

There is much simpler solution. Angle ACB is 90 =› triangles ABC and BOE are similar =› OB/EB=CB/AB =› CB=AB*OB/EB =› CB=2√10*√10/√15=20/√15=4√15/3. Result the same but solution more elegat.

[@GraeTheGreat-zl9us]

What app he using???

[@alamshaikhahmad2415]

5×5=25×4=100sqrooth=10

[@민동익-y3j]

is this for 8grade students?

[@cya3mdirl158]

10:05 not obvious

[@sergovoy]

Корневая труба)))

[@Misha-g3b]

2, -2, 2i, -2i.

[@GUIDORAFAEL2304]

Show!

[@chimaths-class]

Great

[@ivantatarchuk697]

Why OD is the radius of semi circle?

[@cabinetdecuriositestechniq3059]

Yes, why is it the radius ? The entire demonstration is based on this false premise.

[@LukovaMadubo]

A distance from the centre to the circumference of the circle is having a name,it is called RADIUS

[@cabinetdecuriositestechniq3059]

@@LukovaMadubo ...the supposed center of the circle. Nothing shows that O is the center of the circle. Therefore, OD is not the radius.

[@ivantatarchuk697]

@@LukovaMadubo Sorry. I haven't noticed that O is the center.

[@flastkchance5312]

Turkish children are given only 2 minutes to solve this.

[@ΓΕΩΡΓΙΟΣΔΟΥΖΕΝΗΣ-χ9κ]

There is another way of doing it simplified yielding the same resault.

[@mehmeteglen1521]

10 bence bir dakika sürmedi

[@RafaelCardoso-gu6pq]

Mel, o cara que erra essa aí não acerta nem o local de prova

[@wasimahmad-t6c]

10

[@sorourhashemi3249]

❤

[@RK-tf8pq]

Do proper video editing before presenting.

[@wasimahmad-t6c]

5×1.4142135624=7.07×7.07=50×3.14159268=157.07-10×10=57.07÷4=14.2699

[@amritpatel3794]

You took long rout. The angle ACB is 90* , since it lie within semi circle. and EOB is also 90*. So both triangle ABC & EBC are similar triangles. And you are done !!!