Sweden Math Olympiad Geometry Problem | Find the length AD 

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.

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*

Belíssima solução. Obrigado, mestre.

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

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

Thank you.Good explanation

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.

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

Спасибо!

Great one!

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!

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.

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Without pen: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-hU7jERYj04s.html

we have to prove that the triange ABD is similar to the triangle ABC.t hen AD/3 = 6/4. AD - 9/2