Russian Math Olympiad | A Very Nice Geometry Problem
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- Опубликовано: 18 апр 2024
- Russian Math Olympiad | A Very Nice Geometry Problem | Square inside a semicircle
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For the ending I used similar triangles ABC and EBO rather than chord properties. It was simpler to do.
I did this same way 😁
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.
I bounced the square to the right side and applied the same properties as the final part of the solution used in the video.
at 7:40 it simpliest to draw the CA line and take cos(OBE)=cos(ABC)! The solution just appears in front of you!
BC = (4/3)*sqrt(3A)
where A = Area of square OEDF
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.
Deop a perpendicular on X, it will cut at x/2 and calculate X straight away
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)。
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.
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
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.
Also by intersecting chords: DE.DE = CE.EB... this is marginally quicker and a bit less messy.
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
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
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.
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)
That is the same way I went with it.
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
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 !! 🙂
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.
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
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.
Found angle CBA. Then took Cos (CBA) and multiplied it by the diameter of (2*10^.5) = 5.16......
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
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.
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
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.
Exactly. I missed that somehow! Therefore I found it really hard.
@@mickodillon1480 , don't worry. Be happy!
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
You can also join AC and ACB and BOE are similar triangles and take side proportion
Such ingenious thinking!!!!
Good use of Thales theorem.
I did it the same way.
🤙Nice geometry problem bro!...
14+ 4root 5....? Line construction...similarly...diameter...Pythagoras
IF BD=R then AO=OB and OD=R ERROR
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.
is this for 8grade students?
Great
5×5=25×4=100sqrooth=10
|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)
매일 매일 재미있는 수학 영상을 올려주셔서 감사합니다!
I'm korean student
Mel, o cara que erra essa aí não acerta nem o local de prova
10 bence bir dakika sürmedi
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.