The lines SO and QP intersect at point E and the triangle ESQ is a right triangle, hence SE is the diameter of the circle, so SE = 8. According to the Pythagorean theorem EQ = √55, we have cosQES √55/8. Applying the AI Kashi theorem to the triangle EOP, we find x² = 4² + (√55-2)² - 2(4)(√55-2) * √55/8 = 20 - 2√55, so x = √(20 - 2√55)
Complete the circle. Extend SO until it intersects circle at point T. Since ST passes through center O of circle, then OS is a radius and ST is a diameter: ST = 2 * OS = 2 * 4 = 8 Since ∠SQP = 90°, then this angle must be subtended by a diameter with one endpoint at S (ie diameter ST). Therefore, if we extend QP it will also intersect circle at T. Since △QST is a right triangle, we can use Pythagorean Theorem to find QT QT² = ST² − QS² = 8² − 3² = 64 − 9 = 55 QT = √55 Extend radius AO to other side of circle at point C. Diameter AC is divided into segments *AP = 4 − x* and *PC = 4 + x* Chord QT is divided into segments *QP = 2* and *PT = √55 − 2* Using intersecting chord theorem, we get 2 (√55 − 2) = (4 − x) (4 + x) 2√55 − 4 = 16 − x² x² = 20 − 2√55 *x = √(20 − 2√55) ≈ 2.273*
Thank you, Math Booster, for another problem. (This time both easy enough and hard enough for me.) I tackled it this way: QOS is a (4,4,3) triangle . Working out angle OQS: arccos( 4.4 +3.3 -4.4) / 2.4.3 = arccos (3/ 8) = theta {cosine rule} {OQS + OQP = 90 degrees = arccos (0) so sin( OQS + OQP) = (1 - 9/64) + 9/64 =1} sin (OQP) = cos OQS = 3/8 1- 9/64 = 55/64= cos^2(90 -theta) OQP is a (4,2,X) triangle. X^2 = 4.4 +2.2 - 2. 4.2.cos (OQP ) cos OQ P = ( 16+4 - X^2)/16 { cosine rule again, with a twist} cos^2 (OPQ) = (20-x^2)^2 / 256 = 55/64 55 = (20-X^2)(20-X^2)/4 220 = (400 -40X^2 + X^4) X^4 -40 X^2 + (400-220 ) = 0 X^2 = (40 +/- sqrt(1600- 4.1.180) )/2 1600-720 = 880= 4. 220 < 30^2 so the smaller value is correct because X is less than (the radius ) 4. X"2 = 20 - sqrt(220) X= +sqrt(20- sqrt(220) X approx = sqrt(5.17) = 2.27
By connecting point O and point Q, we get ⊿OQS. Since ⊿OQS is an isosceles triangle, if we set ∠QOS＝2θ, then ∠OQS=90°-θ. ∴∠PQO=θ. Applying the law of cosines to ⊿OQS, cos2θ=(OS^2+OQ^2-QS^2)/2×OQ×OS=23/32. Since cos2θ=2(cosθ)^2-1, so, cosθ=√55/8. Applying the law of cosines to ⊿QPO, X^2=PQ^2+OQ^2-2OQ×PQcosθ=20-2√55. ∴X=sqrt(20-2√55)
Completamos el círculo y trazamos la prolongación de AO (hasta punto N), completando el diámetro vertical. Trazamos también la prolongación de QP hasta cortar la circunferencia en el punto M. SM serà diámetro porque el ángulo MQS es de 90°. Por tanto, en MQS la hipotenusa (2R) es = 8 y el cateto menor es = 3. Calculamos el otro cateto que es = raíz de 55. Teorema de cuerdas aplicado a las cuerdas QM y AN, que se cortan en P: AP * PN = MP * PQ (4-X) * (4+X) = 2 * [(raíz de 55)-2] Desarrollando y despejando, llegamos al resultado ~ 2.7 unds Saludos
The company owner X , to cheat Income Tax, deducts advertisement expense. The owner of the advertising channel is Y. Find the relation between X and Y.
If we draw the right angled triangle QSR in the semi circle RS=2r=8 QR=√(8²-3²)=√55 PS=√(2²+3²)=√13 x²=(√13)²+4²-2×√13×4 cos(PSO) PSO=QSR-QSP QSR=α=acos(3/8)=asin((√55)/8) QSP=β=acos(3/√13)=asin(2/√13) x²=13+16-(8√13)cos(α-β) x²=29-(8√13)(cosαcosβ+sinαsinβ) x²=29-8√13{(cos acos3/8)(cos acos3/√13)+(sin asin((√55)/8)(sin asin2/√13) x²=29-8√13{(3/8)(3/√13)+((√55)/8)(2/√13)} x²=29-(9+2√55)=20-2√55 x=√(20-2√55)
the tricky part was the negative root. vnimanye! 10 print "mathbooster-russian math olympiad geometry dec2024":dim x(3),y(3) 20 l1=4:l2=3:l3=2:sw=sqr(l1^2+l2^2+l3^2)/67:l4=sqr(l3^2+l2^2):xp=0:xs=sw:goto 70 30 ys=sqr(l1^2-xs^2):yp=ys+sqr(abs(l4^2-xs^2)):yq=(l1^2-l3^2+yp^2)/2/yp 40 if yq>l1 then return 50 xq=sqr(abs(l1^2-yq^2)):dgu1=(xp-xq)*(xs-xq)/l1^2:dgu2=(yp-yq)*(ys-yq)/l1^2 60 dg=dgu1+dgu2:return 70 gosub 30:if yq>l1 then else 90 80 xs=xs+sw:goto 70 90 xs1=xs:dg1=dg:xs=xs+sw:xs2=xs:gosub 30:if dg1*dg>0 then 90 100 xs=(xs1+xs2)/2:gosub 30:if dg1*dg>0 then xs1=xs else xs2=xs 110 if abs(dg)>1E-10 then 100 120 print "x=";yp: fe=(xq^2+yq^2-l1^2)/l1^2*100:print "der fehler=";fe;"%" 130 x(0)=0:y(0)=0:x(1)=xs:y(1)=ys:x(2)=xq:y(2)=yq:x(3)=0:y(3)=yp:goto 150 140 xbu=x*mass:ybu=y*mass:return 150 masx=1200/l1:masy=850/l1:if masx run in bbc basic sdl and hit ctrl tab to copy from the results window.
The lines SO and QP intersect at point E and the triangle ESQ is a right triangle, hence SE is the diameter of the circle, so SE = 8. According to the Pythagorean theorem EQ = √55, we have cosQES √55/8. Applying the AI Kashi theorem to the triangle EOP, we find x² = 4² + (√55-2)² - 2(4)(√55-2) * √55/8 = 20 - 2√55, so x = √(20 - 2√55)
Complete the circle and by the Intersecting Chords Theorem at P,
(4-X)(4+X)=2*(√55 - 2)
X=√(20-2√55)
Complete the circle. Extend SO until it intersects circle at point T. Since ST passes through center O of circle, then OS is a radius and ST is a diameter:
ST = 2 * OS = 2 * 4 = 8
Since ∠SQP = 90°, then this angle must be subtended by a diameter with one endpoint at S (ie diameter ST). Therefore, if we extend QP it will also intersect circle at T.
Since △QST is a right triangle, we can use Pythagorean Theorem to find QT
QT² = ST² − QS² = 8² − 3² = 64 − 9 = 55
QT = √55
Extend radius AO to other side of circle at point C.
Diameter AC is divided into segments *AP = 4 − x* and *PC = 4 + x*
Chord QT is divided into segments *QP = 2* and *PT = √55 − 2*
Using intersecting chord theorem, we get
2 (√55 − 2) = (4 − x) (4 + x)
2√55 − 4 = 16 − x²
x² = 20 − 2√55
*x = √(20 − 2√55) ≈ 2.273*
Well done. Did it the exact same way. Much easier.
Pytagorean theorem :
(2*4)² = 3² + (y+2)²
y² + 4y + 4 = 8²-3² = 55
y² - 4y -51 = 0. --> y= 5,4162cm
tan α = 3/(2+y) --> α = 22,0243°
Cosine rule :
x² = y² + 4² - 2*4*y*cosα
x = 2,2732 cm ( Solved √ )
Pytagorean theorem :
(2*4)² = 3² + (y+2)²
y² + 4y + 4 = 8²-3² = 55
y² - 4y -51 = 0. --> y= 5,4162cm
Intersecting chords theorem
(R-x)(R+x) = 2.y
R²-x² = 2.y
x² = R² - 2y = 4² - 2*5,4162
x = 2,2732 cm ( Solved √ )
Thank you, Math Booster, for another problem. (This time both easy enough and hard enough for me.)
I tackled it this way:
QOS is a (4,4,3) triangle . Working out angle OQS: arccos( 4.4 +3.3 -4.4) / 2.4.3 = arccos (3/ 8) = theta {cosine rule}
{OQS + OQP = 90 degrees = arccos (0) so sin( OQS + OQP) = (1 - 9/64) + 9/64 =1} sin (OQP) = cos OQS = 3/8 1- 9/64 = 55/64= cos^2(90 -theta)
OQP is a (4,2,X) triangle. X^2 = 4.4 +2.2 - 2. 4.2.cos (OQP ) cos OQ P = ( 16+4 - X^2)/16 { cosine rule again, with a twist} cos^2 (OPQ) = (20-x^2)^2 / 256 = 55/64
55 = (20-X^2)(20-X^2)/4 220 = (400 -40X^2 + X^4) X^4 -40 X^2 + (400-220 ) = 0
X^2 = (40 +/- sqrt(1600- 4.1.180) )/2 1600-720 = 880= 4. 220 < 30^2 so the smaller value is correct because X is less than (the radius ) 4.
X"2 = 20 - sqrt(220) X= +sqrt(20- sqrt(220) X approx = sqrt(5.17) = 2.27
in triangle OQS
cosine rule to find cos(OQS)=3/8
OQS+OQP=90
cos(OQP)=root55/8
also cosine rule in triangle OQP then we can get x
(solved)
By connecting point O and point Q, we get ⊿OQS. Since ⊿OQS is an isosceles triangle, if we set ∠QOS＝2θ, then ∠OQS=90°-θ. ∴∠PQO=θ. Applying the law of cosines to ⊿OQS, cos2θ=(OS^2+OQ^2-QS^2)/2×OQ×OS=23/32.
Since cos2θ=2(cosθ)^2-1, so, cosθ=√55/8.
Applying the law of cosines to ⊿QPO, X^2=PQ^2+OQ^2-2OQ×PQcosθ=20-2√55.
∴X=sqrt(20-2√55)
Completamos el círculo y trazamos la prolongación de AO (hasta punto N), completando el diámetro vertical. Trazamos también la prolongación de QP hasta cortar la circunferencia en el punto M. SM serà diámetro porque el ángulo MQS es de 90°. Por tanto, en MQS la hipotenusa (2R) es = 8 y el cateto menor es = 3. Calculamos el otro cateto que es = raíz de 55.
Teorema de cuerdas aplicado a las cuerdas QM y AN, que se cortan en P:
AP * PN = MP * PQ
(4-X) * (4+X) =
2 * [(raíz de 55)-2]
Desarrollando y despejando, llegamos al resultado ~ 2.7 unds
Saludos
The company owner X , to cheat Income Tax, deducts advertisement expense. The owner of the advertising channel is Y. Find the relation between X and Y.
If we draw the right angled triangle QSR in the semi circle
RS=2r=8
QR=√(8²-3²)=√55
PS=√(2²+3²)=√13
x²=(√13)²+4²-2×√13×4 cos(PSO)
PSO=QSR-QSP
QSR=α=acos(3/8)=asin((√55)/8)
QSP=β=acos(3/√13)=asin(2/√13)
x²=13+16-(8√13)cos(α-β)
x²=29-(8√13)(cosαcosβ+sinαsinβ)
x²=29-8√13{(cos acos3/8)(cos acos3/√13)+(sin asin((√55)/8)(sin asin2/√13)
x²=29-8√13{(3/8)(3/√13)+((√55)/8)(2/√13)}
x²=29-(9+2√55)=20-2√55
x=√(20-2√55)
(2)^2 (3)^2.(4)^2={4+9+16}=29 90°AB/29= =3.3AB (AB ➖ 3AB+3).
So good. Thanks 🎉
Extend SO after completing the circle
A very good geometry problem🎉
X,2x+5=8
2,273
the tricky part was the negative root. vnimanye!
10 print "mathbooster-russian math olympiad geometry dec2024":dim x(3),y(3)
20 l1=4:l2=3:l3=2:sw=sqr(l1^2+l2^2+l3^2)/67:l4=sqr(l3^2+l2^2):xp=0:xs=sw:goto 70
30 ys=sqr(l1^2-xs^2):yp=ys+sqr(abs(l4^2-xs^2)):yq=(l1^2-l3^2+yp^2)/2/yp
40 if yq>l1 then return
50 xq=sqr(abs(l1^2-yq^2)):dgu1=(xp-xq)*(xs-xq)/l1^2:dgu2=(yp-yq)*(ys-yq)/l1^2
60 dg=dgu1+dgu2:return
70 gosub 30:if yq>l1 then else 90
80 xs=xs+sw:goto 70
90 xs1=xs:dg1=dg:xs=xs+sw:xs2=xs:gosub 30:if dg1*dg>0 then 90
100 xs=(xs1+xs2)/2:gosub 30:if dg1*dg>0 then xs1=xs else xs2=xs
110 if abs(dg)>1E-10 then 100
120 print "x=";yp: fe=(xq^2+yq^2-l1^2)/l1^2*100:print "der fehler=";fe;"%"
130 x(0)=0:y(0)=0:x(1)=xs:y(1)=ys:x(2)=xq:y(2)=yq:x(3)=0:y(3)=yp:goto 150
140 xbu=x*mass:ybu=y*mass:return
150 masx=1200/l1:masy=850/l1:if masx
run in bbc basic sdl and hit ctrl tab to copy from the results window.
Very nice
🙏🙏🙏💰💴💸💵