Poland Math Olympiad Problem | A Nice Geometry Challenge
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- Опубліковано 27 лис 2023
- Poland Math Olympiad Problem | A Nice Geometry Challenge
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Mr. Gnome
ஆத்தங்கரை சண்முகம்
Εκπαιδευτήρια Καντά
堤修一
Sidnei Medeiros Vicente
Sambasivam Sathyamoorthy
Muhtar Samedov
Gerard van Bemmel
Khusi Wala
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Sam
Fatih Ekşioğlu
I started off the same way you did with Pythagorean theorem in △OEC
r² + x² = (r+1)²
r² + x² = r² + 2r + 1
x² = 2r + 1
But then, instead of using similar triangles, I used Pythagorean theorem in △ABC
3² + (2r+1)² = (x+3)²
9 + (x²)² = x² + 6x + 9
x⁴ − x² − 6x = 0
x (x³ − x − 6) = 0
Ignoring solution x = 0 (which would give r = −1/2), we get
x³ − x − 6 = 0 → (same equation we get by using similar triangles)
(x − 2) (x² + 2x + 3) = 0
The quadratic factor has no real solutions, we get
x = 2
2r + 1 = x² = 4
2r = 3
r = 3/2
Your English is so bad
Were you learn to write "r"?
You can also reflect the triangle about BC to form a triangle with an inscribed circle.
Then the formula A=Pr/2 can be applied, where A is the area of the large triangle, P is its perimeter, and r is the required radius.
We have that 6(2r+1)/2=r(2sqrt((2r+1)^2+9)+6)/2, which ultimately boils down to the cubic equation 2r^3+r^2-9=0, yielding the solution r=3/2.
I solved this as follows:
Let B' be a point with AB'=3+r on the extended line of AB.
Let C' be a point with BC//B'C' on the extended line of AC.
The upper half of the inscribed circle of △AB'C' is equivalent to the given semi-circle.
AB'=r+3
B'C'=(2r+1)(r+3)/3
AC'^2=AB'^2+B'C'^2
AB'+B'C'-AC'=2r
You can get a cubic equation of r from these equations.
2r^3+r^2-9=0
⇔(2r-3)(r^2+2r+3)=0
⇒r=3/2(∵r∈R+)
i have applied the thales theorem. that's a sophisticated nested calculation:
10 l1=1:l2=3:nu=29:dim x(2),y(2):sw=(l1+l2)/100:xs=sw:nl=sqr(l1^2+l2^2):goto 120
20 z1=ys*l1:z2=l2*(xs-l1):dfu1=(z1+z2)/2/(l2-ys)/nl
30 dfu2=(xs^2+ys^2)/2/xs/nl:df=dfu1-dfu2:return
40 ys=sw:gosub 20
50 df1=df:ys1=ys:ys=ys+sw:if ys>l2 then stop
60 ys2=ys:gosub 20:if df1*df>0 then 50
70 ys=(ys1+ys2)/2:gosub 20:if df1*df>0 then ys1=ys else ys2=ys
80 if abs(df)>1E-10 then 70
90 r=(xs^2+ys^2)/2/xs:rt=(r+l1)/2:xmt=r+rt
100 dgu1=(xs-xmt)^2/nl^2:dgu2=ys^2/nl^2:dgu3=rt^2/nl^2:dg=dgu1+dgu2-dgu3
110 return
120 gosub 40
130 xs1=xs:dg1=dg:xs=xs+sw:rem if xs>l1 then stop
140 xs2=xs:gosub 40:if dg1*dg>0 then 130
150 xs=(xs1+xs2)/2:gosub 40:if dg1*dg>0 then xs1=xs else xs2=xs
160 if abs(dg)>1E-10 then 150
170 print r:mass=1000/nl:x(1)=r*2+l1:y(1)=0:x(2)=0:y(2)=l2:goto 190
180 xb=x*mass:yb=y*mass:return
190 xba=0:yba=0:for a=1 to 3:ia=a:if ia=3 then ia=0
200 x=x(ia):y=y(ia):gosub 180:xbn=xb:ybn=yb:goto 220
210 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
220 gosub 210:next a:xm=r:ym=0:ru=r:goto 260
230 x=xm+ru:y=0:gosub 180:xba=xb:yba=yb:for a=1 to nu:wa=a/nu*pi
240 dx=ru*cos(wa):x=xm+dx:dy=ru*sin(wa):y=ym+dy:gosub 180:xbn=xb:ybn=yb
250 gosub 210:next a:return
260 gcol 7:gosub 230:xm=xmt:ru=rt:gcol 9:gosub 230
1.5
>
run in bbc basic sdl and hit ctrl tab to copy from the results window
3-4-5-triple 🙂
R=3/2.
8:46 is where I have a problem. Its trial and error here.
Exactly!
To make this look less like a lucky guess, let's go down a slightly different path. So far we have x³ = 6 + x. Now continue:
x³ - x = 6
x(x² - 1) = 6
x(x+1)(x-1) = 6
So we need three numbers that differ by 1 (basically an arithmetic sequence) whose product is 6. Clearly, the numbers have to be 1, 2, 3. Now, you can argue it's lucky this was so easy to see, but then any result that wasn't easy to see would require finding the roots of a cubic, and surely the creators of the test wouldn't expect you to do THAT, right? 😊
Bravo
If you write sin of angle C in two ways , you get 4R^4 + 4r^3 + R^2 - 18R - 9 = 0 and you can find that R= 3/2 is a root. Then devide the polinom by (R - 1.5) and you get 4R^3 +10R^2 +16R +6 and this polinom is increasing for all R [based on its derivative] , and for R=0 it is 6 so it has one negative root.
How can you find that R = 3/2 is a root? That is the crux of the problem. I know it is a root, and the solution, but how to get it except by guesswork?
Are you aware of the theorm regarding the possible rational roots ?@@gibbogle
@@user-mx8sj1nc6v No.
Square, not esquiee
Which olympiad is that? Task is suspiciously similar to the one in ua-cam.com/video/K9VKCfYPHiE/v-deo.htmlsi=uJNXuXeRv0my6W0L. If you are taken somebody's else task it would be fair to leave a reference to this video.
You should have stopped when you got x = 2, which was a lucky guess.