Japan Math Olympiad | A Very Nice Geometry Problem
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Assign the square a side length of "2" and ...
Take a glance at :
1) The right triangle with R1 as a leg and point E as a vertex.
2) The right triangle with R2 as a leg and point E as a vertex.
Notice that they are similar and from this we have that R2/R1 = (1 + sqrt(5))/2.
This follows because R2/(1 - R2) = (1 - R1)/R1 and since R2 = 2/(3 + sqrt(5)) from which we can find R1.
Let's set AE=EB=a=1. So The radius of the inscribed circle of △ADE is R2 = (3-√5)/2.
Suppose the center and tangent point of the right circle are P and N, and the center and tangent point of the left circle are Q and M.
Then we find that ∠PEQ = 90° and △QME～ △PEN. So we know that R2/(a-R2) = (a-R1)/R1.
Finally we solve R1 = (√5-1)/2 due to R2 = (3-√5)/2. Thus, R1/R2 = (3-√5)/(√5-1) = (√5+1)/2.
there are 2 solutions:
either find the intersections of angle bisectors or a repeated calculation:
10 print "mathbooster-japan math olympiad":dim x(1,3),y(1,3)
20 l1=1:xg1=0:yg1=l1:xg2=l1/2:yg2=0:sw=l1/10
30 a11=xg2-xg1:a12=yg2-yg1:a21=yg1-yg2:a22=xg2-xg1:a231=yg1*(xg2-xg1):a232=xg1*(yg1-yg2)
40 a23=a231+a232:r1=sw:goto 140
50 xp=r1:yp=xp:gosub 70:dgu1=(xp-xl)^2/l1^2:dgu2=(yp-yl)^2/l1^2:dg=dgu1+dgu2
60 dgu3=r1^2/l1^2:dg=dgu1+dgu2-dgu3:goto 130
70 a131=xp*(xg2-xg1):a132=yp*(yg2-yg1):a13=a131+a132
80 ngl1=a12*a21:ngl2=a22*a11
90 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end
100 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2
110 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2
120 xl=zx/ngl:yl=zy/ngl:rem print "x=";xl;"y=";yl
130 return
140 gosub 50
150 dg1=dg:r11=r1:r1=r1+sw:r12=r1:gosub 50:if dg1*dg>0 then 150
160 r1=(r11+r12)/2:gosub 50:if dg1*dg>0 then r11=r1 else r12=r1
170 if abs(dg)>1E-10 then 160
180 xm1=xp:ym1=yp:r2=sw:goto 210
190 xp=l1-r2:yp=r2:gosub 70:dgu1=(xp-xl)^2/l1^2:dgu2=(yp-yl)^2/l1^2:dg=dgu1+dgu2
200 dgu3=r2^2/l1^2:dg=dgu1+dgu2-dgu3:return
210 gosub 190
220 dg1=dg:r21=r2:r2=r2+sw:r22=r2:gosub 190:if dg1*dg>0 then 220
230 r2=(r21+r22)/2:gosub 190:if dg1*dg>0 then r21=r2 else r22=r2
240 if abs(dg)>1E-10 then 230
250 print "das verhaeltnis r1/r2=";r1/r2:xm2=xp:ym2=yp
260 x(0,0)=0:y(0,0)=0:x(0,1)=l1/4:y(0,1)=0:x(0,2)=l1/2:y(0,2)=0:x(0,3)=0:y(0,3)=l1
270 x(1,0)=l1/2:y(1,0)=0:x(1,1)=l1:y(1,1)=0:x(1,2)=l1:y(1,2)=l1:x(1,3)=0:y(1,3)=l1
280 mass=1000/l1:goto 300
290 xbu=x*mass:ybu=y*mass:return
300 for a=0 to 1:x=x(a,0):y=y(a,0):gosub 290:xba=xbu:yba=ybu:for b=1 to 4:c=b:if c=4 then c=0
310 x=x(a,c):y=y(a,c):gosub 290:xbn=xbu:ybn=ybu:goto 330
320 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
330 gosub 320:next b:next a:gcol 12:x=xm1:y=ym1:gosub 290:circle xbu,ybu,r1*mass
340 x=xm2:y=ym2:gosub 290:circle xbu,ybu,r2*mass
350
mathbooster-japan math olympiad
das verhaeltnis r1/r2=0.618033989
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run in bbc basic sdl and hit ctrl tab to copy from the results window. you may add"@zoom%=1.4*@zoom%" at the start.
@ 3:32 why do not you use the known formula for the inscribed circle in a triangle
r = 2 Areas/ a+b+c ?
Yes. It is simple and easy to establish : Area = 0.5*a*b = 0.5*R*(a+b+c)
you will find the same result with substitution of a, b, c in R = a*b / (a+b+c)
@@WahranRai yes, but the formuls in the video is different.
@@ludmilaivanova1603 It is special formula in case of right triangle.
I will recommend to forget it and use the general one as you said
Solution by trigonometry. Label the center of the small circle as point F, the center of large circle as point G, the point of tangency of small circle with AE as point H and point of tangency of large circle with BE as point J. Let
360°ABCDE/2 =180°ABCDE 2^90 2^3^30 2^3^5^6 2^3^5^3^2 1^15^1^3^2 1^1^3^2 3^2 (ABCDE ➖ 3ABCDE+2)
Let s be the side length of square ABCD. Let O be the center of the larger circle (with racius R₁), let P be the cemter of the smaller circle (with radius R₂), let M be the point of tangency between circle O and DE, let N be the point of tangency between circle O and AB, and let T be the point of tangency between circle O and BC.
As ME and NE are tangents to circle O that meet at E, ME = NE = x. As NB = R₁ and BE = s/2, x = (s/2)-R₁.
Mirror square ABCD about AB, forming square ABFG and draw EF. As E is the midpoint of AB and AD = CB = BF = GA = s, DF will be a straight line.
As MF and TF are both tangents to circle O that meet at F, MF = TF = y. As BF = s and BT = R₁, y = R₁+s. As MF = ME+ EF, y also equals x+EF.
Triangle ∆FBE:
FB² + BE² = EF²
s² + (s/2)² = EF²
EF² = s² + s²/4 = 5s²/4
EF = √(5s²/4) = √5s/2
y = x + EF
R₁+s = (s/2)-R₁ + √5s/2
2R₁= √5s/2 - s/2 = s(√5-1)/2
R₁ = s(√5-1)/4
Draw JK, where J is a point on BC and K is a point on DE where JK is perpendicular to BC and is tangent to circle O opposite N, at point L. This forms a right triangle ∆FJK.
As ∠DAE = ∠FJK = 90° and ∠EDA = ∠KFJ as alternate interior angles, triangles ∆DAE and ∆FJK are similar. Additionally, circles O and P are inscribed in each triangle in the same manner, so R₁ and R₂ are in the same ratio as the sides of the respective triangles.
Triangle ∆DAE:
R₁/R₂ = FJ/DA
R₁/R₂ = (s+2R₁)/s
R₁/R₂ = (s+2s(√5-1)/4)/s
R₁/R₂ = 1 + (√5-1)/2
R₁/R₂ = (2+√5-1)/2 = (1+√5)/2 = φ
Le relazioni sono l-R1+(l/2)-R1=√5l/2...(l-2R1+R2)^2-(l-R2)^2=(l-R2)^2....dalla prima risulta l=4R1/(3-√5) che metto nella seconda,divido per R1R2,e risulta R2/R1=(1+√5)/2...NB..HO SCAMBIATO R1 CON R2
Golden answer
The answer is -1/2(-1-sqrt(5)). I actually think that you may have done a similar problem on your channel. I could be wrong.
Apparently this shows that R1/R2 equals R2/R1.
Or equivalently, ((1+sqrt(5))/2).
@@herbcruz4697 Yes. I wished that I simplified that further. I appreciate that clarification. I hope that I do not sound like an idiot.
R1=a-R2😂