Russian Math Olympiad | A Very Nice Geometry Problem

[@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

[@mohamadmehdimotamed4485]

از تشابه دو مثلث OEB و ACB به حالت دو زاویه برابر زودتر به جواب میرسیم .(زاویه B در هر دو مشترک ،زاویه O و C هم برابر 90 درجه.

[@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!

[@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.

[@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.

[@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 .

[@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.

[@MorgKev]

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

[@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 !! 🙂

[@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. 😉

[@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

[@감나빗-26]

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

[@اقرء]

Tanks for watching

[@magdyeldajany9973]

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

[@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.

[@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.

[@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.

[@bakrantz]

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

[@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)

[@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).

[@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

[@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.

[@maxgladkikh2326]

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

[@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)

[@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.

[@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)

[@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.

[@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.

[@kaistmechanic]

If you use the similarity condition of a triangle, it's much easier to find the answer

[@dlspark7965]

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

[@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 √ )

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال (cosB)^2=5/4r^2 X^2=(2r)^2 (1+2cos2B) = 2(2r)^2(cosB)^2 X^2= 10

[@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*

[@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 .

[@tamirerez2547]

Very nice solution 👍

[@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 → BC = 4√15/3

[@plamenpenchev262]

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

[@user-gn4mq5cs6e]

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

[@whoskiana]

you can also solve this by similarity, can't you? Triangle EOB ~ ACB (ACB=EOB=90° and EBO is common), from there you can find BC

[@InventPeace1]

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

[@gaylespencer6188]

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

[@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

[@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

[@mg.1044]

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

[@TheCodeFather254]

Why not create the line AC and use trigonometry to solve for CB?

[@GraeTheGreat-zl9us]

What app he using???

[@احمدالجوهر-ظ4ت]

OB×AB=CB×EB ACB right angled

[@TyyylerDurden]

I found BC through the similarity between the right triangle CAB (it is right because its hype is the circle's diameter) and the triangle OEB. 20/root 15

[@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 !

[@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.

[@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.

[@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

[@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.

[@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.

[@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

[@retarody9542]

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

[@michallesz2]

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

[@Cricketdoctor_1999]

🤙Nice geometry problem bro!...

[@sakibalmahmud686]

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

[@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.

[@shoebmd428]

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

[@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

[@cya3mdirl158]

10:05 not obvious

[@감나빗-26]

is this for 8grade students?

[@wiwa1962]

|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)

[@MYldrm]

Five ore pipe?

[@himadrikhanra7463]

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

[@just_isaac15]

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

[@inventionssimpleexpression7694]

Trigonometri kulnarak çözmek daha pratik ve şık

[@SGuerra]

Uau! Que questão bonita. Eu encontrei uma solução um pouco diferente. Parabéns pela escolha!!! Brasil - setembro de 2024.

[@pepe_valitor487]

&.Square OEDF: 5=(side)² > side=√5 side=OE=ED=DF=FO=√5 __ __ __ &.∆ODFrect: OF=DF=√5 ; OD=r=rad __ __ __ (Teor.Pithag.) OD²=OF²+DF² __ OD²=(√5)²+(√5)²=2(√5)²=2•5=10 __ OD=√10=r __ __ &.∆ACBrect: AB=2r=2√10 ; CB=x __ __ &.∆EOBrect: OE=√5 ; OB=√10 __ __ __ (T.Pythag.) EB²=OE²+OB² __ EB²=(√5)²+(√10)²=5+10=15 __ EB=√15 &.∆ACB~∆EOB. (similar∆, propiedad AAA) __ __ __ __ AB/CB=EB/OB __ __ __ __ AB=2√10 ; CB=x ; EB=√15 ; OB=√10 (2•√10)/x=(√15)/√10 2•(√10)•√10=x•√15 > 2•(√10)²=x•√15 2•10=x•√15 > 20=x•√15 > x=20/√15 x=(20•√15)/(√15•√15)=(20•√15)/15 _______________________ x=(4•√15)/3=~5.16cm✔️ 👍 _______________________

[@Dauooria]

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

[@alamshaikhahmad2415]

5×5=25×4=100sqrooth=10

[@vrcfncpdci]

A good problemi for mr. perrelman the russian genius

[@chimaths-class]

Great

[@Misha-g3b]

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

[@ΓΕΩΡΓΙΟΣΔΟΥΖΕΝΗΣ-χ9κ]

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

[@RafaelCardoso-gu6pq]

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

[@sergovoy]

Корневая труба)))

[@flastkchance5312]

Turkish children are given only 2 minutes to solve this.

[@GUIDORAFAEL2304]

Show!

[@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.

[@RK-tf8pq]

Do proper video editing before presenting.

[@sorourhashemi3249]

❤

[@mehmeteglen1521]

10 bence bir dakika sürmedi

[@wasimahmad-t6c]

10

[@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 !!!