France Math Olympiad | A Very Nice Geometry Problem

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[michaeldoerr]

The answer is 4*sqrt(2). This is probably the third or fourth problems where the Law od Cosines was applied twice and probably the third or fourth porblem that required knowledge of the curcumcenter. And apparently the Law of Cosines is always applicable to finding the radius of the circumcenter. I could be wrong.

[quigonkenny]

As ∠ABP = ∠PBC = θ, then the arca subtended by ∠ABP and ∠PBC, are equal. Thus AP = PC = x.

By the law of cosines:

Triangle ∆ABP:
AP² = AB² + BP² - 2AB(BP)cos(θ)
x² = 4² + 10² - (2)4(10)cos(θ)
x² = 16 + 100 - 80cos(θ)
x² = 116 - 80cos(θ) --- [1]

Triangle ∆PBC:
PC² = PB² + BC² - PB(BC)cos(θ)
x² = 10² + 11² - (2)10(11)cos(θ)
x² = 100 + 121 - 220cos(θ)
x² = 221 - 220cos(θ) ---- [2]

116 - 80cos(θ) = 221 - 220cos(θ) <-- [1] = [2]
140cos(θ) = 221 - 116 = 105
cos(θ) = 105/140 = 3/4

x² = 116 - 80(3/4) <-- [1]
x² = 116 - 60 = 56
x = √56 = 2√14

Let O be the center of the circle. By the double angle theorem, as point B is on the circumference and O is at the center, as ∠PBC = θ, ∠POC = 2θ.

By the law of cosines:

Triangle ∆POC:
PC² = OP² + OC² - 2OP(OC)cos(2θ)
x² = r² + r² - (2)r(r)cos(2θ)
56 = 2r² - 2r²cos(2θ)
2r²(1-cos(2θ)) = 56
r² = 28/(1-cos(2θ))
r² = 28/(1-(2cos²(θ)-1))
r² = 28/(2-2(3/4)²)
r² = 28/(2-2(9/16))
r² = 28/(2-9/8) = 28/(7/8)
r² = 28(8/7) = 32
r = √32 = 4√2 units

[solimana-soli]

I have found cos theta with Ptolemy theorem on ABCP and considering that triangle APC is isosceles and CA = 2x cos theta
4x + 11x = 10*2x*cos theta => cos theta = 3/4

[nishagupta]

Join the segment to form cyclic quad then use law of cosines to find the length of the segment to be √56 then realize that both joint segments are equal then use Ptolemy to find the other diagonal as 3√14 then join it to the radius then find the value of cos २Theta, notice then that diagonal 2 will subtend angle 4theta at tne centre using value of cos 2 theta find value of cos 4theta again apply cosine rule and you are done as the answer is 4√2

[Alessandro-]

I didn' t understand why AP = PC ?

[giuseppemalaguti]

Io ho eguagliato la formula r=abc/4A...in questo modo risulta una soluzione per cosθ=3/4.e cosθ=5/22+√(47, 5)/11...perciò calcolo r(cosθ=3/4)=4√2..r(θ=31, 37..)=6, 7587..

[oscarcastaneda]

My Math was with me as I saw that the corresponding segments would be equal and would relate to the radius at double the given inscribed angles. From this I was able to determine sin(theta) and cos(theta) which led to Radius as 4sqrt(2).

[RealQinnMalloryu]

(11)^2=121 (10)^2=100 (4)^2, =16 {121+100+16}=237 360°ABCP/237=1.123 1^1.10^10^23 10^10^23^1 10^10^1^1 2^52^5 1^1^2^1 2^1 (ABCP ➖ 2ABCP+1).

[zdrastvutye]

et voilà! quelquechose digne pour une grande nation, si quelqu'un
connait les 4 mousquetaires:
10 print "math booster-france math olympiad":dim x(3), y(3)
20 130
30
40
50
60 ngl1=a12*a21:ngl2=a22*a11
70 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end
80
90
100 xl=zx/ngl:yl=zy/ngl:rem print "x=";xl;"y=";yl
110
120 dg=dg-1:return
130 gosub 30
140 dg1=dg:w1=w:w=w+sw:w2=w:gosub 30:if dg1*dg>0 then 140
150 w=(w2+w1)/2:gosub 30:if dg1*dg>0 then w1=w else w2=w
160 if abs(dg)>1E-10 then 150 else xm=xl:ym=yl
170 print "der winkel theta=";deg(w), "R=";r
180
190 if masx<masy then mass=masx else mass=masy
200 goto 220
210
220 x=x(0):y=y(0):gosub 210:xba=xbu:yba=ybu:for a=1 to 3
230 x=x(a):y=y(a):gosub 210:xbn=xbu:ybn=ybu:goto 250
240 line xba, yba, xbn, ybn:return
250 gosub 240:next a:x=xm:y=ym:gosub 210:circle xbu, ybu, r*mass
math booster-france math olympiad
der winkel theta=41.4096221 R=5.65685425
>
executez avec bbc basic sdl et frappez ctrl tab pour copier len contenu du fenetre "resultats"

[murdock]

ABP = PBC = δ; AB = 4; BC = 11; BP = 10; AP = CP = k; AO = BO = CO = PO = r = ?
k^2 = 116 - 80cos⁡(δ) = 221 - 220cos⁡(δ) → cos⁡(δ) = 3/4 →
sin⁡(δ) = √7/4 → cos⁡(2δ) = 1/8 → k^2 = 56 = 2r^2(1 - cos⁡(2δ)) → r = 4√2