Sweden Math Olympiad Geometry Problem

Sweden Math Olympiad Geometry Problem | Find the length AD

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Sweden Math Olympiad Geometry Problem | Find the length AD
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Пікірлер: 28

@MarieAnne. 11 ай бұрын

Nice solution. Simpler than mine, which went as follows: Let AD = x, CD = y. We need to find x. Using Law of Sines in △BCD, we get: sin C / 3 = sin 2θ / y = 2 sin θ cos θ / y sin C = 6 sin θ cos θ / y Using Law of Sines in △ABC, we get: sin C / 6 = sin θ / 4 sin C = 3 sin θ / 2 So we get: 6 sin θ cos θ / y = 3 sin θ / 2 cos θ = y/4 Using Law of Cosines in △ABD, we get: 3^2 = 6^2 + x² − 2(6)(x) cos θ 9 = 36 + x² − 12x(y/4) 0 = 27 + x² − 3xy 3xy = x² + 27 y = (x²+27)/(3x) Using Law of Cosines in △ABC, we get: 4^2 = 6^2 + (x+y)² − 2(6)(x+y) cos θ 16 = 36 + (x+y)² − 12(x+y) (y/4) (x+y)² + 20 − 3y(x+y) = 0 Substitute in y = (x²+27)/(3x) → x+y = (4x²+27)/(3x) ((4x²+27)/(3x))² + 20 − 3((x²+27)/(3x))((4x²+27)/(3x)) (4x²+27)²/(9x²) + 20 − 3(x²+27)(4x²+27)/(9x²) Multiply through by 9x² (x²+27)² + 180x² − 3(x²+27)(4x²+27) = 0 4x⁴ − 9x² − 1458 = 0 (x² + 18) (4x² − 81) Since x is a real positive number, then: x² = 81/4 *x = 9/2*

@zdrastvutye 10 ай бұрын

this is a sophisticated nested calculation! 10 l1=3:l2=4:l3=6:sw=.01:goto 110 20 dgu1=l3/l2*sin(w):dgu2=sin(pi-3*w-w2) 30 dg=dgu1-dgu2 :return 40 w2=sw:gosub 20 50 dg1=dg:w21=w2:w2=w2+sw:if w2>pi then stop 60 w22=w2:gosub 20:if dg1*dg>0 then 50 70 w2=(w21+w22)/2:gosub 20:if dg1*dg>0 then w21=w2 else w22=w2 80 if abs(dg)>1E-10 then 70 90 wu=pi-2*w-(pi-3*w-w2):dfu1=l2/l1*sin(pi-3*w-w2):dfu2=sin(wu) 100 df=dfu1-dfu2 :return 110 w=sw:gosub 40 120 df1=df:wu1=w:w=w+sw:if w>pi then stop 130 wu2=w:gosub 40:if df1*df>0 then 120 140 w=(wu1+wu2)/2:gosub 40:if df1*df>0 then wu1=w else wu2=w 150 if abs(df)>1E-10 then 140 160 print w:lx=l1*sin(w2)/sin(w):print"x="; lx:rem probe 170 l3r=lx*cos(w)+l1*cos(w2):print l3r:fe=(1-l3/l3r)*100:print "der fehler ist=";fe;"%" 180 dim x(2),y(2):mass=1500/(l1+l2+l3):goto 200 190 xb=x*mass:yb=y*mass:return 200 x(0)=-l3*cos(w2+2*w):y(0)=0:x(1)=x(0)+l2:y(1)=0:x(2)=0:y(2)=l3*sin(w2+2*w) 210 x=x(0):y=y(0):gosub 190:xba=xb:yba=yb:xbe=xb:ybe=yb:for a=1 to 3:ia=a:if ia=3 then ia=0 220 x=x(ia):y=y(ia):gosub 190:xbn=xb:ybn=yb:goto 240 230 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 240 gosub 230:next a:dx=l1*cos(2*w):dy=l1*sin(2*w):xf=x(0)+dx:yf=y(0)+dy 250 xba=xbe:yba=ybe:x=xf:y=yf:gosub 190:xbn=xb:ybn=yb:gosub 230:x=x(2):y=y(2) 260 gosub 190:xbn=xb:ybn=yb:gcol 9:gosub 230 0.50536051 x=4.5 6 der fehler ist=-1.51406981E-8% > run in bbc basic sdl and hit ctrl tab to copy. the graphics shows the real angles

@اممدنحمظ Жыл бұрын

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

@alexbayan8302 4 ай бұрын

Triangle EBC similar to BAC. So BE=6y; Triangle BED similar to AEB so 2=BE:DE=AE:BE;AE=2*BE=12y; AC=12y+4y=16y In similar triangles EBC and BAC, 4:16y=4y:4 ; 64*y^2=16; y*y=1/4 ; y=1/2 AD=AE-DE=12y-3y=9y=9/2 The basic angle bisector theorem is used to get the ratio 3y and 4y and then nothing else is used after that except the similarity arguments.

@ROCCOANDROXY Жыл бұрын

Let angle(BDC) = lambda and draw a perpendicular from B to AC intersecting at E implies BE = 3 * sin(lambda) = 6 * sin(theta) implies cos^2(lambda) = 1 - sin^2((lambda) = 4 * cos^2(lambda) - 3 and 3 * sin(lambda) = 4 * sin(2 * theta + lambda) implies 3 = 2 * (2 * cos(theta) * sqrt(4 * cos^2(theta) - 3) + 4 * cos^2(theta) - 2) implies (7 - 8 * cos^2(theta))^2 = 16 * cos^2(theta) * (4 * cos^2(theta) - 3) implies 49 = 64 * cos^2(theta) implies cos^2(theta) = 49/64 implies cos(theta) = 7/8 implies 9 = x^2 + 36 - 12 * x * (7/8) using law of cosines on triangle(ABD) implies 2 * x^2 - 21 * x + 54 = 0 implies (2 * x - 9) * (x - 6) = 0 implies x = 9/2, x = 6. x = 6 implies Triangle(BAD) is isosceles triangle implies m(angle(ABD)) = m(angle(ADB)) = 180 - lambda implies lambda = 90 + theta/2 implies sin(90 + theta/2) = cos(theta/2) = 2 * sin(theta) implies 1 - cos(theta) = 8 * sin^2(theta) implies 8 * cos^2(theta) - cos(theta) - 7 = 0 implies (8 * cos(theta) + 7) * (cos(theta) - 1) = 0 implies cos(theta) = 1 or cos(theta) = -7/8 implies theta = 0 or theta > 90 implies 2 * theta > 180, therefore we drop x = 6 and x = 9/2.

@zianiera Жыл бұрын

Thank you.Good explanation

@murdock5537 Жыл бұрын

Nice and awesome, many thanks, Sir, this is great! ∆ABC → AB = 6; BC = 4; AD = 3; AC = AD + CD = x + CD; CAB = φ; DBC = 2φ → EBC = DBE = φ; BE = b; x = ? 3/4 = DE/CE → DE = (3/4)CE → CE = 4z → DE = 3z → b^2 = 12 - 12z^2 ∆ABC ≈ ∆BCE → 4/6 = 4z/b → 4b = 24z → b = 6z → b^2 = 36z^2 = 12 - 12z^2 → z = 1/2 → 7z = 7/2 → 3/4 = 6/(x + 7/2) → x = 9/2 btw: AC = x + 7z = 8 49/4 = 9 + 16 - 2(3)(4)cos⁡(2φ) = 25 - 24cos⁡(2φ) → cos⁡(2φ) = 17/32 → sin⁡(2φ) = 7√15/32 → cos⁡(φ) = 7/8 → sin⁡(φ) = √15/8 ABC = ϑ → cos⁡(ϑ) = -1/4

@luigipirandello5919 Жыл бұрын

Belíssima solução. Obrigado, mestre.

@vkr122 Жыл бұрын

Uz BC pagarinājuma atliekkam punktu X tā , ka XAB=BAC=ʘ , tad XAB ~ CBD , AX/AC=3x/4x , XB=4 , AB - bisektrise AB=√(3x*4x-3*4), x=2, XA=6, AC=8, XAB~CBD k=2, DC=XC/2=3.5 AD=8-3.5=4.5!

@solomou146 Жыл бұрын

Καλημέρα σας. Επειδή στα Ελληνικά βιβλία δεν αναφέρεται ο τύπος AD^2 = AB*AC-BD*CD, θα μπορούσατε να μου υποδείξετε μία απόδειξη σε αυτόν και αν είναι "επώνυμο" θεώρημα, το όνομά του; Ευχαριστώ.

@1ciricola Жыл бұрын

I agree. We are owed an explanation about this unfamiliar equation. It was invoked in two KZbin videos [explaining the same problem] with no explanation. I searched the Internet with no success.