A beautiful piece of Geometry!

This is a beautiful and exciting math olympiad question from the first round of 2022 math olympiads in Turkey. Solution of the problem requires the use of;
- Trigonometry,
- Pythagorean theorem,
- Cosine theorem,
- Intersecting secants theorem,
- Square of tangent theorem,
- Midpoint theorem on right-angled triangles,
- Algebra.

Chapters:
(00:00) - Question
(00:34) - Solution Method 1
(03:07) - Solution Method 2

#geometry
#matholympiadquestion
#matholympiadpreparation
#maths
#mathematics
#triangle
#rightangletriangle

[じーちゃんねる-vn]

AngleCAD=CBE=θ AngleOED=ODE=α sinθ=1/3 cosθ=2√2/3 from sine rule for CDE sin(θ+α)=(2/3)cosα sinα=cosα/(2√2) cosα=(2/3)√2

[じーちゃんねる-vn]

Suppose x(1/√2) plane D(0, 0) A(0, 2√2) B(-1, 0) C(1, 0) AC:y=-√2(x-1) BE:y=(1/(2√2))(x+1) E(7/9, (4/9)√2) O(x, y) (9-14x)/4=2√2y=x+1 x=5/18 y=23/(36√2) r=√(x^2+y^2)=3/(4√2) real r=3/4

[tellerhwang]

△ADC~△EDG(AA)
→3sqrt2:2r=4:sqrt2
→r=3/4😊

[幕天席地-wc]

The method 1 can be simplified. Besides connecting DE, also connect DG(G is the other point BE intersects the circle). Triangle EDG and BEC are similar. so GE:DE = BC:BE. namely 2*r : sqrt(2) = 2*sqrt(2) : 8/3, r = 3/4.

[じーちゃんねる-vn]

F is the middle of EC, EDF is the same as CDF DE=√2 angleBAD=CBE=DEB=θ sinθ=1/3 cosθ=2√2/3 DE=2rcosθ=√2 r=3/4

[edsznyter]

In the first solution, when you have the isosceles triangle inside the circle, just draw the height through O; the two subtriangles are similar to ABD, so just do the ratios instead of all the trig. Ah, same idea as user-ii5vr3um9r..

[dfx]

Hocam türk müsünüz? Ben muhtesem uclu yapip cemberin merkezinden muhtesem uclu olan kirise dik indirdim ve benzerlikten yaptim, hos soru

[zdrastvutye]

this works without trigonometry, there you go:
10
30
40 sw=(l1+l2)/1E2:rem r und xm, ym des kreises berechnen
50 90
60
70
80 dg=dgu1+dgu2-dgu3-dgu4:return
90 k=sw:gosub 60
100 k1=k:dg1=dg:k=k+sw:if k>10*l1 then stop
110 k2=k:gosub 60:if dg1*dg>0 then 100
120 k=(k1+k2)/2:gosub 60:if dg1*dg>0 then k1=k else k2=k
130 if abs(dg)>1E-10 then 120
140 r=sqr((xm-l2)^2+ym^2):print "der radius=";r
150 dim x(2), y(2):la=l1:lb=la:lc=2*l2:gosub 30:x(0)=0:y(0)=0
160 mass=6E2/n:goto 180
170 xb=x*mass:yb=y*mass:return
180 xba=0:yba=0:for a=1 to 3:ia=a:if ia=3 then ia=0
190 x=x(ia):y=y(ia):gosub 170:xbn=xb:ybn=yb:goto 210
200 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
210 gosub 200:next a:x=xs3:y=ys3:gosub 170:xbn=xb:ybn=yb:gcol 5:gosub 200
220 gcol6:x=xm:y=ym:gosub 170:circle xb, yb, r*mass
230
der radius=0.75
>
run in bbc basic sdl and hit ctrl tab to copy from the results window. (sometimes the geometry in the clip is not true, so better calculate it)

[Михаил-дхз]

(√(2)÷cos(acos((sin(acos(1
÷3))×2×√(2))÷(2×√(2))))÷2
=
0.75

[幕天席地-wc]

Another solution: O is the centre of the circle. F is the other point DC intersects the circle. H is the midpoint of DF. Triangle BHO and BEC are similar. Suppose r is radius of the circle. OD is r, OH is (8/3 - r)/3 and DH is (8/3 - r)*2*sqrt(2)/3 - sqrt(2). OD^2 = OH^2 + DH^2.