Wow, neat solution. Here is how I did it, which is so totally different: Let AD = d Join A and C. Since BD = CD = 15, then angles subtended by arcs BC and CD will be equal. Therefore, ∠BAC = ∠CAD = θ, and ∠BAD = 2θ Since AB is a diameter, then △ABC has right angle at C and sin θ = sin(∠BAC) = BC/AB = 15/d Join B and D to form △ABD. Since AB is a diameter, then △ABD has right angle at D and cos(∠BAD) = AD/AB = 7/d But we can also calculate as follows: cos(∠BAD) = cos 2θ = 1 − 2sin²θ = 1 − 2(15/d)² = (d²−450)/d² Now we equate both values of cos(∠BAD) (d²−450)/d² = 7/d d² − 450 = 7d d² − 7d − 450 = 0 (d − 25) (d + 18) = 0 Since d is a diameter, it must be positive: *d = 25*
Alternative short solution. Let E be the point of intersection of the lines AD and BC. Since DC=BC and AD are the diameter, AC is the bisector and height of the BAE triangle. Therefore, AE=AB=d, BC=CE=15. From the equality EA * ED = EB * EC we get the equation d*(d-7) = 30 * 15. It follows that d=25. The root d=-18 does not satisfy.
There was once a very similar problem. First year of USAMTS (1989-1990), round 4, question 1: A hexagon is inscribed in a circle of radius r. Find r if two sides of the hexagon are 7 units long, while the other four sides are 20 units long.
Another way would be by exploiting the symmetries as follows. Define M to be the pont that halfs the half circle in the video. Flip the triangle ACD along the side AC, to get D'. Because of symmetry the point D' is part of the circle with radius r and center O. Also because of symmetry, the line CD' is parallel to line AB and the line OM is perpendicular to line CD' and cutting it exactly in half. Use the coordinate system with O=(0; 0), B=(r; 0), M=(0; r). The given circle is described by x^2 + y^2 = r^2. Then point C has an x value of 7/2 and is an intersection of the half circle above and a second circle with center B and radius 15 (with x^2-2xr+r^2 + y^2 = 225). ==> x^2 + y^2 - r^2 = (x-r)^2 + y^2 - 15^2 x^2 + y^2 - r^2 = x^2-2xr+r^2 + y^2 - 225 0 = (2r)^2 - 2(2r)x - 450 0 = (2r)^2 - 2(2r)(7/2) + (7/2)^2 - (49/4) - 1800/4 0 = (2r - 7/2)^2 - (43/2)^2 0 = (2r - 50/2)(2r + 36/2) 2r = 25 or 2r = -18 | d = 2r >= 0 ==> d = 25
We have cyclic quadrilateral so angles DAB + DCB = 180 Triangle ADB is right triangle (inscribed angle based on semicircle) Cosine rule twice (first time in the triangle ADB with angle DAB second time in triangle DCB with angle DCB) cos(alpha) from basic trigonometry (SOA,CAH,TOA) and i have got polynomial equation of degree three with three real roots but two of them are negative
Angle BOC is x. Angle AOD is 180-2x. Radius is r. In triangle BOC with generalized Pithagora: r^2 + r^2 - 2*r*r*cos(x) = 15^2 In triangle AOD: r^2 + r^2 - 2*r*r*cos(180-2x) = 7^2 cos(180-2x) = - cos2x =2cos^2(x) - 1 Solve the sistem of the 2 equation, you find cos (x) and radius.
Problem can be made MUCH SIMPLER if you first re-arrange the 3 lines, putting the 7 one between the two 15 long ones, without changing the problem. That makes construction symmetrical about vertical line through circle center (bisects the 7 line), & the 7 line is parallel to the bottom diameter line. This vertical line (center to intersection of the 7 long line) is part of right triangle hypotenuse is R, short leg is 7/2 (because of symmetry). This line's length squared is = R^2 - (7.2)^2. There is another right triangle formed: hypotenuse 15, short leg R - (7/2), & long leg same as other triangle, R^2 - (7.2)^2. Applying poth theorem to this triangle gives equation 2R^2 - 7R -225 = 0. Solution to this is R = 12.5 (neg solution rejected), giving D = 25.
Nice but then you have to JUSTIFY why your configuration and the original one give the same radius ... it is not hard to explain but it has to be done 😊
@@ericvuillemey2135Justification is: a given chord subtends given angle regardless where it is in the circle. The problem here is to have the 3 subtended angles sum to exactly a half-turn. That summation process is commutative, just like summation of numbers. It doesn't matter the order.
Cosine theorem: 7² = R² + R² - 2R² cos α 7² = 2R² - 2R² cos α 7² = 2R² (1 - cos α) Cosine theorem: 15² = R² + R² - 2R² cos β 15² = 2R² - 2R² cos β 15² = 2R² (1 - cos β) Supplementary angles: 180° = α + 2 β Put these formulas in an Excel worksheet and will obtain : R = 12,5 cm D = 25 cm. (Solved √ )
Join DB and CA as diagonals of the cyclic quad. AD=7,CD =15,BC=15 and AB=d say. angle ACB= angle ADB=90 deg. since AB=d =diameter of cyclic quad. Rule:- 'the sum of the products of the opposite sides equals the product of the diagonals.' for a cyclic quad. Rule (1) say. Consider triangle ADB, d^2=AD^2+BD^2 Pythagoras BD= sqrt(d^2-49).......................(1) Consider triangle ACB, d^2=AC^2+CB^2 Pythagoras CA= sqrt(d^2-225)....................(2) The product of the opposite sides; (1)DC*d=15d............................(3) (2DA*CB=7*15=105.................(4) Sum of the products of opposite sides =15d+105....................(5) Product of diagonals = BD*CA=sqrt(d^2-49)*sqrt(d^2-225).........................(6) Rule(1) means:- 15d+105=sqrt(d^2-49)*sqrt(d^2-225) square each side, (15d+105)^2=(d^2-49)(d^2-225) 225d^2+3150d+11025=d^4-274d^2+11025 d^4-499d^2-3150d=0................................................................(7) This factors to; d(d-25)(d+7)(d+18)=0 d=0, d=25, d= -7 d= -18. The only useful solution is d=25 units and that is the answer. Assisted by Wolfram Alpha for the factoring. Thanks for the problem.
Method of factoring without using Wolfram Alpha; If you look at triangle ADB it looks as it it is a 25,24,7 triangle Therefore, d=25 units could be one of the four roots of d^4-499d^2-3150d=0, d^4-499d^2-3150d=0 d(d^3-0*d^2-499d-3150)=0, d=0 is one root but this does not fit into the cyclic quad. Consider the cubic (d^3-0*d^2- 499d-3150)=0 -3150 factors into +/-1,+/-7,+/-18 and +/-25. dividing the cubic(d^3-0*d^2- 499d-3150)=0 by (d-25) gives (d^2+25d+126)=0 d^2+25d+126=(d+7)(d+18)=0 The four roots are therefore, d=0, d=25, d= -7, d=-18 d=25 units is the only practical answer.
Here a short solution based on the known trigonometric identity cos(2x) = 1 - 2 * sin(x)^2: We can extract the following 3 equations from the task: sin(a) = 7/d sin(b) = 15/d a + 2*b = pi/2 (note: both a and b are the half angles, so a + 2b sums up to 90° = pi/2) Using those and the above identity we get sin(a) = sin(pi/2 - 2*b) = cos(2*b) = 1 - 2 * sin(b)^2, and therefor 7/d = 1 - 2 * (15/d)^2, which can be reshaped to: d^2 - 7d - 450 = 0. Solving this quadratic equation we find: d = (7 +/- 43)/2, and the only positive solution is therefor d = 25.
AD^2 + DB^2 = d^2 => 7^2 + DB^2 = d^2 AC^2 + CB^2 = d^2 => 15^2 + AC^2 = d^2 Observation of lengths in the diagram indicates that the Pythagorean triples which apply here are (7, 24, 25) and (15, 20, 25). Thus d = 25.
@@dwschiuAlso, seeing that side with length 7 doesn't mean a 7-24-25 triangle will come into play. It turns out there is a 7-24-25 triangle here, but the composer of this problem undoubtedly did that and had no particular reason to do so. Now, finding the criteria for which the diameter, the two adjacent sides, and the remaining side of the quadrilateral are all integers strikes me as an interesting question. It turns out this is true when a² + 8b² is a perfect square, where b is the length of the two adjacent sides and a the length of the remaining side. So the integer solutions to the equation a² + 8b² = c² fit the bill - it's similar to the Pythagorean Theorem but not quite - giving a diameter equal to (a + c)/2. One integral solution that I find interesting is with a = 7 and b = 3, giving c = 11 and diameter = 9. So the lengths of all four sides are small integers, but you won't find any Pythagorean triples here. 😊
Brute force approach: Note that AO, BO, CO and DO are radii, call their length R. Drop perpendiculars from O to AD, call the intersection E, O to CD, call the intersection F, and O to BC, call the intersection G. Note that ΔOBG, ΔOCG, ΔOCF, and ΔODF are congruent.
Draw AC and BD and use sine rule in triangle ACB and ADB. ∠DAC = θ ∠CAB = θ ∠DCA = 90-2θ ∠DBA = 90-2θ ∠ADB = 90 ∠ACB = 90 from △ACB using sine-law 15/sinθ = d from △ADB using sine-law 7/sin(90-2θ)=d 7/cos2θ =d so 15/sinθ = 7/cos2θ cos2θ/sinθ = 7/15 (1-2sin^2θ)/sinθ = 7/15 (As cos2θ =1-2sin^2θ) if x=sinθ (1-2x^2)/x= = 7/15 solving this Quadratic equation it sinθ will be 0.6 from △ACB using sine-law 15/sinθ = d d=15/0.6 = 25
once r has been chosen, the coordinates of c and d can be calculated 10 l1=15:l2=15:l3=7:dim x(3),y(3):sl=l1+l2+l3:sw=sl/100:yc=1:nu=55:r=sw:goto 70 20 xd=l3^2/2/r:yd=l3^2-xd^2:if yd1E-10 then 110 130 print "r=";r:p=sw:goto 150 140 dg=(l1^p+l2^2+l3^p-(2*r)^p)/sl^p:return 150 gosub 140 160 p1=p:dg1=dg:p=p+sw:if p>10*l1 then stop 170 p2=p:gosub 140:if dg1*dg>0 then 160 180 p=(p1+p2)/2:gosub 140:if dg1*dg>0 then p1=p else p2=p 190 if abs(dg)>1E-10 then 180 200 print l1;"^";p;"+";l2;"^";p;"+"; l3;"^";p;"="; (2*r);"^";p 210 x(0)=0:y(0)=0:x(1)=r*2:y(1)=0:x(2)=xc:y(2)=yc:x(3)=xd:y(3)=yd 220 print xc,"%",yc,"%",xd,"%",yd 230 mass=500/r:goto 250 240 xb=x*mass:yb=y*mass:return 250 x=0:y=0:gosub 240:xba=xb:yba=yb:for a=1 to 4:ia=a:if ia=4 then ia=0 260 x=x(ia):y=y(ia):gosub 240:xbn=xb:ybn=yb:goto 280 270 line xba,yba,xbn,ybn:return 280 gosub 270:xba=xbn:yba=ybn:next a:x=2*r:y=0:gosub 240:xba=xb:yba=yb 290 for a=1 to nu+1:wa=a/nu*pi:dx=r*cos(wa):dy=r*sin(wa):x=r+dx:y=dy 300 gosub 240:xbn=xb:ybn=yb:gosub 270:xba=xbn:yba=ybn:next a r=12.5 15^1.88190314+15^1.88190314+7^1.88190314=25^1.88190314 16% 12% 1.96% 6.72 > run in bbc basic sdl and hit ctrl tab to copy from the results window
I like to solve everything with trigonometry. Obviously 7/(2*sin(x/2)) = 15/(2*sin((Pi-x)/4)) => sin(x/4) = √2/10 and sin(x/2) = 7/25 => d = 2* (7/(2*7/25)) = 25.
I heard about the concept of “Indian code” (Indians were paid for the amount of code and they wrote as much code as possible). But this is the first time I’ve seen Indian mathematics... No offense.
Join the diagonals AC and BD. Now apply Ptolemy's Theorem to cyclic quadrilateral ABCD 15d + 15 X 7 = V(d^2 - 15^2) X V(d^2 - 7^2) Square both sides 225(d+7)^2 = (d^2-225)(d+7)(d-7) Divide both sides by (d+7) (d^2-225)(d-7) = 225(d+7) d^3 - 7d^2 - 225d + 225 X 7 = 225d + 225 X7 d^3 - 7d^2 - 450d = 0 Divide by d d^2 - 7d - 450 =0 (d+18)(d - 25) = 0 d cannot be -18 and so d= 25 Sumith Peiris Moratuwa Sri Lanka
Looking into the reasons why my previous solution was wrong, I found this other solution: We trace the radii that join the center of the semicircle with the extremes of the three chords and also those perpendicular to them. We obtain 4 congruent triangles whose side measures 15/2 and 2 congruent triangles whose side measures 7/2. Using a little trigonometry we can write 15/2 = r*sinx 7/2 = r*siny now we know that 4x + 2y = 180° then y = 90° - 2x for which 7/2 = r*sin(90° - 2x) using the formulas of the associated arcs sin(90° - 2x)=cos2x and cos2x = 1 - 2sin²x then we solve the system of equations by treating sinx as one of the unknowns (r,sinx) 15/2=r*sinx 7/2=r*(1-2sin²x) that gives 2r²-7r-255=0 that gives r=25/2
OE is connecting middles of sides to triangle DAB,so OE = DA/2 = 7/2. OC=r and CE=CO-EO= r - (7/2). COMPLETE CIRCLE and CO intercepts in point F, so FE= r +(7/2) . Aplying Intercepting Chords Theorem into point E for chords DB and CF, we have DE*EB = CE*EF so since DE=EB DE^2 = CE*EF so DE^2 = ( r - (7/2) )*( r +( 7/2) ) (1) Also from PYTH.THEOR to DEC right triangle DE^2=15^2- ( r - (7/2))^2 (2) From equalities (1),(2) we have ( r - (7/2) )*( r +( 7/2) ) = 15^2- ( r - (7/2))^2 so r^2 -(49/4) = 225 - ( r - (7/2))^2 AND 4*r^2 - 49 = 900 - 4 r ^2 +28*r - 49 so 8*r^2 - 28*r - 900 = 0 . (Dividing by 4) 2*r^2 - 7*r - 225 = 0 . D=49+1800=1849 = 43^2 . Hence r1= (7+43) /4 = 50/4 = 25/2 ACCEPT and r2=(7 - 43)/4 = - 9 REJECT. Finally d= 2*r = 2* 25/2 = 25.
Let O be the center of the semicircle, at the midpoint of diameter AB. Draw OC and OD. As OB = OC = OD = r and BC = CD = 15, ∆BOC and ∆COD are congruent isosceles triangles. As OD = OA, ∆DOA is also an isosceles triangle. Let ∠BOC = x. As ∠COD = ∠BOD, and AB is the diameter, ∠DOA = 180-2x. By the law of cosines we have two equations: cos(x) = (r²+r²-15²)/2r² cos(x) = (2r²-225)/2r² cos(180-2x) = (r²+r²-7²)/2r² cos(2x) = (49-2r²)/2r² 2cos²(x) - 1 = (49-2r²)/2r² 2((2r²-225)/2r²)² - 1 = (49-2r²)/2r² 2((u-225)/u)² - 1 = (49-u)/u
O be the centre of this semi circle Join radii OB, OC, OD. In quadrilateral OBCD DC = BC & OD = OB Hereby quadrilateral OBCD is a kite Its duagonal CO perpendicularly bisects BD at P. Again ∆ DAB is similar to ∆ POB PO / AD = BO/ AB = 1/2 PO = AD/2 and CP = r - AD/2 Again BO^2 - OP^2 =BP^2 =BC^2 -CP^2 BC^2 - r^2 = CP^2 - OP^2 = ( CP + OP) (CP - OP) = r ( r - OP - OP) = r ( r - AD) Hereby 2 r ^2 - r AD = BC^2 Herein. 2 r ^2 - 7 r - 225 = 0 2 r^2 - 25 r + 18 r -225 = 0 (2 r - 25)( r + 9) = 0 Duameter of semi circle 2 r = 25
I got as far as arcsin(3.5/r)x2+arcsin(7.5/r)x4=180. I could only complete by trial and error as I had no idea how to isolate r. I was hoping to find out that there was a way to do that.
Sin alterar las premisas del problema, se puede reordenar el esquema inicial de forma que quede una figura simétrica con la cuerda de 7 unidades de longitud, horizontal y centrada en la parte alta del semicírculo 》El triángulo de la derecha (de base "r" y lados "r" y 15) se compone de dos triángulos rectángulos con el cateto vertical común, hipotenusas "r" y 15 y bases (7/2) y (r -7/2) 》 r^2 - (7/2)^2 = 15^2 - (r - 7/2)^2 》 r=25/2 》 d=2r =2(25/2) =25=d Gracias y saludos.
Hey I was just wondering did you make use of one of the circle theorems? I ask this bc that might be the proof of why OD and OC are radii. I could be wrong.
Νομίζω ότι δώσατε την πιο κατάλληλη λύση στο πρόβλημα αυτό. Η δική μου 1η σκέψη ήταν το θεώρημα του Πτολεμαίου (ACxBD=ADxBC+ABxCD) αλλά η αλγεβρική της επίλυση είναι δύσκολη από πλευράς πράξεων.
I did this reasoning, but the result is a little different. If the three chords of the semicircle were of the same length, it would be half the size of a regular hexagon. So I divided the overall length (15+15+7)/3=12.333... In a regular hexagon inscribed in a circle, the side is equal to the radius. What's wrong?
Too algebraic, can solve it more geometrically. Extend ad and bc, meet at e, you got 2 similar triangles, note the mid line, rest is pce of cake.......
I do not have a good English. Let be x and y the diagonals of the cyclic quadrilateral. So: d^2-49=x^2 d^2-225=y^2. But as the quadrilateral is cyclic xy=15*(7+d) ...(xy)^2=15^2*(d+7)^2 (d^2-49)*(d^2-225)=15^2*(d+7)^2 (d-7)*(d^2-225)=225*(d+7) d^3-7d^2-450d=0 as d0 d^2-7d-450=0 d=25 or d=-18(not good) So d=25.
I passed through obscene amounts of trigonometry ( 15 = 2 r sin α/2 , 7 = 2 r sin (90° - α) and so on) but anyway I came through with the right solution😊 (very inelegant,but it worked).
Solution without angles. Use "diagonals product" (e*f) theorem for cyclic quadrilaterals: e*f = a*c + b*d e*f = 15*2r + 7*15 Combine with two right triangle equations having 2r as their hypoteneuse: e^2 = (2r)^2 - 15^2 f^2 = (2r)^2 - 7^2 Using some algebra, a quartic equation in r was developed, and solved using an online solver. r = 12.5. If Math Olympiad doesn't allow computational tools, it's growing obsolete as a forum for advanced problem-solving.
i made a mistake by proportioning the angles using the cord length. it is not correct. the three angles (along the centre) are proportioning by sine threta/2 vs the cord length. cord length = 2 r sin threta/2. using this relationship we can calculate the 3 angles and the r and hence the dia. the r = 12.5, dia = 25. thanks.
Al ojo Traza el segmento BD y Traza el segmento AC Entonces se forma 90° en D y en C Por arco, segmentos iguales (15) arcos iguales. => digamos angA = 2w Y ang B= w + €, de tal manera que w apunta a 15 y € apunta a 7 Entonces en triangulo ADB angA + € = 90° 2w +€ = 90° Es decir por razones complementarias Sen€ = Cos 2w Además Por triangulos rectángulos Sen€ = 7/d Senw= 15/d De Sen€ = Cos 2w 7/d = 1- 2sen(w)^2 7/d = 1- 2(15/d)^2 Resolviendo d^2 - 7d + 450 = 0
Assuming that the drawing is in scale, I measure length "d" with a school ruler, then I calculate the proportion according to the scale, and it gives me approximately 25 cm d = 25 cm (Solved √ )
Wow, neat solution. Here is how I did it, which is so totally different:
Let AD = d
Join A and C.
Since BD = CD = 15, then angles subtended by arcs BC and CD will be equal.
Therefore, ∠BAC = ∠CAD = θ, and ∠BAD = 2θ
Since AB is a diameter, then △ABC has right angle at C and
sin θ = sin(∠BAC) = BC/AB = 15/d
Join B and D to form △ABD.
Since AB is a diameter, then △ABD has right angle at D and
cos(∠BAD) = AD/AB = 7/d
But we can also calculate as follows:
cos(∠BAD) = cos 2θ = 1 − 2sin²θ = 1 − 2(15/d)² = (d²−450)/d²
Now we equate both values of cos(∠BAD)
(d²−450)/d² = 7/d
d² − 450 = 7d
d² − 7d − 450 = 0
(d − 25) (d + 18) = 0
Since d is a diameter, it must be positive:
*d = 25*
BD = x, AC = y, x*2 + 49 = d*2, y*2 + 225 = d*2, (Ptolemy's theorem) 15d + 105 = xy, (d*2 - 49)(d*2 -225) = (15d + 105)*2 (d > 0), solving by the usual algebra methods, d = 25.
Alternative short solution.
Let E be the point of intersection of the lines AD and BC. Since DC=BC and AD are the diameter, AC is the bisector and height of the BAE triangle. Therefore, AE=AB=d, BC=CE=15. From the equality EA * ED = EB * EC we get the equation d*(d-7) = 30 * 15. It follows that d=25. The root d=-18 does not satisfy.
تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .
There was once a very similar problem. First year of USAMTS (1989-1990), round 4, question 1:
A hexagon is inscribed in a circle of radius r. Find r if two sides of the hexagon are 7 units long, while the other four sides are 20 units long.
you are right
Another way would be by exploiting the symmetries as follows.
Define M to be the pont that halfs the half circle in the video.
Flip the triangle ACD along the side AC, to get D'.
Because of symmetry the point D' is part of the circle with radius r and center O.
Also because of symmetry, the line CD' is parallel to line AB
and the line OM is perpendicular to line CD' and cutting it exactly in half.
Use the coordinate system with O=(0; 0), B=(r; 0), M=(0; r).
The given circle is described by x^2 + y^2 = r^2.
Then point C has an x value of 7/2 and is an intersection of the half circle above and a second circle with center B and radius 15 (with x^2-2xr+r^2 + y^2 = 225).
==> x^2 + y^2 - r^2 = (x-r)^2 + y^2 - 15^2
x^2 + y^2 - r^2 = x^2-2xr+r^2 + y^2 - 225
0 = (2r)^2 - 2(2r)x - 450
0 = (2r)^2 - 2(2r)(7/2) + (7/2)^2 - (49/4) - 1800/4
0 = (2r - 7/2)^2 - (43/2)^2
0 = (2r - 50/2)(2r + 36/2)
2r = 25 or 2r = -18 | d = 2r >= 0
==> d = 25
We have cyclic quadrilateral so angles DAB + DCB = 180
Triangle ADB is right triangle (inscribed angle based on semicircle)
Cosine rule twice (first time in the triangle ADB with angle DAB second time in triangle DCB with angle DCB)
cos(alpha) from basic trigonometry (SOA,CAH,TOA)
and i have got polynomial equation of degree three with three real roots but two of them are negative
Angle BOC is x.
Angle AOD is 180-2x.
Radius is r.
In triangle BOC with generalized Pithagora:
r^2 + r^2 - 2*r*r*cos(x) = 15^2
In triangle AOD:
r^2 + r^2 - 2*r*r*cos(180-2x) = 7^2
cos(180-2x) = - cos2x =2cos^2(x) - 1
Solve the sistem of the 2 equation, you find cos (x) and radius.
I used Ptolemy's theorem to arrive at d³ - 499d - 3150 = 0
Solving gives d = -18, -7 and 25, so d = 25 as the other 2 options are -ve.
Problem can be made MUCH SIMPLER if you first re-arrange the 3 lines, putting the 7 one between the two 15 long ones, without changing the problem. That makes construction symmetrical about vertical line through circle center (bisects the 7 line), & the 7 line is parallel to the bottom diameter line. This vertical line (center to intersection of the 7 long line) is part of right triangle hypotenuse is R, short leg is 7/2 (because of symmetry). This line's length squared is = R^2 - (7.2)^2.
There is another right triangle formed: hypotenuse 15, short leg R - (7/2), & long leg same as other triangle, R^2 - (7.2)^2. Applying poth theorem to this triangle gives equation 2R^2 - 7R -225 = 0. Solution to this is R = 12.5 (neg solution rejected), giving D = 25.
Cut the half circle along OC. Rotate the small fan CBO from OB to OA.
Nice but then you have to JUSTIFY why your configuration and the original one give the same radius ... it is not hard to explain but it has to be done 😊
@@ericvuillemey2135Justification is: a given chord subtends given angle regardless where it is in the circle. The problem here is to have the 3 subtended angles sum to exactly a half-turn. That summation process is commutative, just like summation of numbers. It doesn't matter the order.
That's what I had too 👍😊
Cosine theorem:
7² = R² + R² - 2R² cos α
7² = 2R² - 2R² cos α
7² = 2R² (1 - cos α)
Cosine theorem:
15² = R² + R² - 2R² cos β
15² = 2R² - 2R² cos β
15² = 2R² (1 - cos β)
Supplementary angles:
180° = α + 2 β
Put these formulas in an Excel worksheet and will obtain :
R = 12,5 cm
D = 25 cm. (Solved √ )
Join DB and CA as diagonals of the cyclic quad.
AD=7,CD =15,BC=15 and AB=d say.
angle ACB= angle ADB=90 deg. since AB=d =diameter of cyclic quad.
Rule:- 'the sum of the products of the opposite sides equals the product of the diagonals.' for a cyclic quad. Rule (1) say.
Consider triangle ADB,
d^2=AD^2+BD^2 Pythagoras
BD= sqrt(d^2-49).......................(1)
Consider triangle ACB,
d^2=AC^2+CB^2 Pythagoras
CA= sqrt(d^2-225)....................(2)
The product of the opposite sides;
(1)DC*d=15d............................(3)
(2DA*CB=7*15=105.................(4)
Sum of the products of opposite sides =15d+105....................(5)
Product of diagonals = BD*CA=sqrt(d^2-49)*sqrt(d^2-225).........................(6)
Rule(1) means:-
15d+105=sqrt(d^2-49)*sqrt(d^2-225)
square each side,
(15d+105)^2=(d^2-49)(d^2-225)
225d^2+3150d+11025=d^4-274d^2+11025
d^4-499d^2-3150d=0................................................................(7)
This factors to;
d(d-25)(d+7)(d+18)=0
d=0,
d=25,
d= -7
d= -18.
The only useful solution is d=25 units and that is the answer.
Assisted by Wolfram Alpha for the factoring.
Thanks for the problem.
Method of factoring without using Wolfram Alpha;
If you look at triangle ADB it looks as it it is a 25,24,7 triangle
Therefore, d=25 units could be one of the four roots of d^4-499d^2-3150d=0,
d^4-499d^2-3150d=0
d(d^3-0*d^2-499d-3150)=0,
d=0 is one root but this does not fit into the cyclic quad.
Consider the cubic (d^3-0*d^2- 499d-3150)=0
-3150 factors into +/-1,+/-7,+/-18 and +/-25.
dividing the cubic(d^3-0*d^2- 499d-3150)=0 by (d-25) gives (d^2+25d+126)=0
d^2+25d+126=(d+7)(d+18)=0
The four roots are therefore,
d=0,
d=25,
d= -7,
d=-18
d=25 units is the only practical answer.
Here a short solution based on the known trigonometric identity cos(2x) = 1 - 2 * sin(x)^2:
We can extract the following 3 equations from the task:
sin(a) = 7/d
sin(b) = 15/d
a + 2*b = pi/2 (note: both a and b are the half angles, so a + 2b sums up to 90° = pi/2)
Using those and the above identity we get sin(a) = sin(pi/2 - 2*b) = cos(2*b) = 1 - 2 * sin(b)^2, and therefor 7/d = 1 - 2 * (15/d)^2, which can be reshaped to:
d^2 - 7d - 450 = 0.
Solving this quadratic equation we find: d = (7 +/- 43)/2, and the only positive solution is therefor d = 25.
Love these geometrical problems
Very instructive solutions.
I wonder if there is an alternative solution..
AD^2 + DB^2 = d^2 => 7^2 + DB^2 = d^2
AC^2 + CB^2 = d^2 => 15^2 + AC^2 = d^2
Observation of lengths in the diagram indicates that the Pythagorean triples which apply here are (7, 24, 25) and (15, 20, 25). Thus d = 25.
Sulit ya klo maen kira2 gini
That would be a nice solution if you knew beforehand that d and the measure of the other segments were natural numbers. That is not always the case.
@@dwschiuAlso, seeing that side with length 7 doesn't mean a 7-24-25 triangle will come into play. It turns out there is a 7-24-25 triangle here, but the composer of this problem undoubtedly did that and had no particular reason to do so.
Now, finding the criteria for which the diameter, the two adjacent sides, and the remaining side of the quadrilateral are all integers strikes me as an interesting question. It turns out this is true when a² + 8b² is a perfect square, where b is the length of the two adjacent sides and a the length of the remaining side. So the integer solutions to the equation a² + 8b² = c² fit the bill - it's similar to the Pythagorean Theorem but not quite - giving a diameter equal to (a + c)/2. One integral solution that I find interesting is with a = 7 and b = 3, giving c = 11 and diameter = 9. So the lengths of all four sides are small integers, but you won't find any Pythagorean triples here. 😊
Brute force approach: Note that AO, BO, CO and DO are radii, call their length R. Drop perpendiculars from O to AD, call the intersection E, O to CD, call the intersection F, and O to BC, call the intersection G. Note that ΔOBG, ΔOCG, ΔOCF, and ΔODF are congruent.
😊
I did the same, but I don't prefer trial and error.
Draw AC and BD and use sine rule in triangle ACB and ADB.
∠DAC = θ
∠CAB = θ
∠DCA = 90-2θ
∠DBA = 90-2θ
∠ADB = 90
∠ACB = 90
from △ACB using sine-law
15/sinθ = d
from △ADB using sine-law
7/sin(90-2θ)=d
7/cos2θ =d
so 15/sinθ = 7/cos2θ
cos2θ/sinθ = 7/15
(1-2sin^2θ)/sinθ = 7/15 (As cos2θ =1-2sin^2θ)
if x=sinθ
(1-2x^2)/x= = 7/15
solving this Quadratic equation it sinθ will be 0.6
from △ACB using sine-law
15/sinθ = d
d=15/0.6 = 25
once r has been chosen, the coordinates of c and d can be calculated
10 l1=15:l2=15:l3=7:dim x(3),y(3):sl=l1+l2+l3:sw=sl/100:yc=1:nu=55:r=sw:goto 70
20 xd=l3^2/2/r:yd=l3^2-xd^2:if yd1E-10 then 110
130 print "r=";r:p=sw:goto 150
140 dg=(l1^p+l2^2+l3^p-(2*r)^p)/sl^p:return
150 gosub 140
160 p1=p:dg1=dg:p=p+sw:if p>10*l1 then stop
170 p2=p:gosub 140:if dg1*dg>0 then 160
180 p=(p1+p2)/2:gosub 140:if dg1*dg>0 then p1=p else p2=p
190 if abs(dg)>1E-10 then 180
200 print l1;"^";p;"+";l2;"^";p;"+"; l3;"^";p;"="; (2*r);"^";p
210 x(0)=0:y(0)=0:x(1)=r*2:y(1)=0:x(2)=xc:y(2)=yc:x(3)=xd:y(3)=yd
220 print xc,"%",yc,"%",xd,"%",yd
230 mass=500/r:goto 250
240 xb=x*mass:yb=y*mass:return
250 x=0:y=0:gosub 240:xba=xb:yba=yb:for a=1 to 4:ia=a:if ia=4 then ia=0
260 x=x(ia):y=y(ia):gosub 240:xbn=xb:ybn=yb:goto 280
270 line xba,yba,xbn,ybn:return
280 gosub 270:xba=xbn:yba=ybn:next a:x=2*r:y=0:gosub 240:xba=xb:yba=yb
290 for a=1 to nu+1:wa=a/nu*pi:dx=r*cos(wa):dy=r*sin(wa):x=r+dx:y=dy
300 gosub 240:xbn=xb:ybn=yb:gosub 270:xba=xbn:yba=ybn:next a
r=12.5
15^1.88190314+15^1.88190314+7^1.88190314=25^1.88190314
16% 12% 1.96% 6.72
>
run in bbc basic sdl and hit ctrl tab to copy from the results window
I like to solve everything with trigonometry. Obviously 7/(2*sin(x/2)) = 15/(2*sin((Pi-x)/4)) => sin(x/4) = √2/10 and sin(x/2) = 7/25 => d = 2* (7/(2*7/25)) = 25.
There was no need for proving similarity (8.24). EO joins mid points of two sides of a triangle .hence it will be half of 7= 3.5.( midpoint theorem)
I heard about the concept of “Indian code” (Indians were paid for the amount of code and they wrote as much code as possible). But this is the first time I’ve seen Indian mathematics... No offense.
Join the diagonals AC and BD.
Now apply Ptolemy's Theorem to cyclic quadrilateral ABCD
15d + 15 X 7 = V(d^2 - 15^2) X V(d^2 - 7^2)
Square both sides
225(d+7)^2 = (d^2-225)(d+7)(d-7)
Divide both sides by (d+7)
(d^2-225)(d-7) = 225(d+7)
d^3 - 7d^2 - 225d + 225 X 7 = 225d + 225 X7
d^3 - 7d^2 - 450d = 0
Divide by d
d^2 - 7d - 450 =0
(d+18)(d - 25) = 0
d cannot be -18 and so d= 25
Sumith Peiris
Moratuwa
Sri Lanka
Great
Teşekkürler.
Thank you for supporting this channel 😊
Looking into the reasons why my previous solution was wrong, I found this other solution:
We trace the radii that join the center of the semicircle with the extremes of the three chords and also those perpendicular to them. We obtain 4 congruent triangles whose side measures 15/2 and 2 congruent triangles whose side measures 7/2. Using a little trigonometry we can write
15/2 = r*sinx
7/2 = r*siny
now we know that
4x + 2y = 180° then
y = 90° - 2x
for which
7/2 = r*sin(90° - 2x)
using the formulas of the associated arcs
sin(90° - 2x)=cos2x and cos2x = 1 - 2sin²x
then we solve the system of equations by treating sinx as one of the unknowns (r,sinx)
15/2=r*sinx
7/2=r*(1-2sin²x) that gives
2r²-7r-255=0 that gives r=25/2
OE is connecting middles of sides to triangle DAB,so OE = DA/2 = 7/2. OC=r and CE=CO-EO= r - (7/2).
COMPLETE CIRCLE and CO intercepts in point F, so FE= r +(7/2) . Aplying Intercepting Chords Theorem into point E for chords DB and CF, we have DE*EB = CE*EF so since DE=EB DE^2 = CE*EF so DE^2 = ( r - (7/2) )*( r +( 7/2) ) (1)
Also from PYTH.THEOR to DEC right triangle DE^2=15^2- ( r - (7/2))^2 (2)
From equalities (1),(2) we have ( r - (7/2) )*( r +( 7/2) ) = 15^2- ( r - (7/2))^2 so r^2 -(49/4) = 225 - ( r - (7/2))^2 AND 4*r^2 - 49 = 900 - 4 r ^2 +28*r - 49 so 8*r^2 - 28*r - 900 = 0 .
(Dividing by 4) 2*r^2 - 7*r - 225 = 0 . D=49+1800=1849 = 43^2 . Hence r1= (7+43) /4 = 50/4 = 25/2 ACCEPT and r2=(7 - 43)/4 = - 9 REJECT. Finally d= 2*r = 2* 25/2 = 25.
Let O be the center of the semicircle, at the midpoint of diameter AB. Draw OC and OD. As OB = OC = OD = r and BC = CD = 15, ∆BOC and ∆COD are congruent isosceles triangles. As OD = OA, ∆DOA is also an isosceles triangle.
Let ∠BOC = x. As ∠COD = ∠BOD, and AB is the diameter, ∠DOA = 180-2x. By the law of cosines we have two equations:
cos(x) = (r²+r²-15²)/2r²
cos(x) = (2r²-225)/2r²
cos(180-2x) = (r²+r²-7²)/2r²
cos(2x) = (49-2r²)/2r²
2cos²(x) - 1 = (49-2r²)/2r²
2((2r²-225)/2r²)² - 1 = (49-2r²)/2r²
2((u-225)/u)² - 1 = (49-u)/u
Beautiful problem. Thank you Sir.
sin(BDA) = sin(φ) = 1 → AB = 25 → ∆ABD = pyth. triple (7-24-25)
O be the centre of this semi circle
Join radii OB, OC, OD.
In quadrilateral OBCD
DC = BC & OD = OB
Hereby quadrilateral OBCD is a kite
Its duagonal CO perpendicularly bisects BD at P.
Again ∆ DAB is similar to ∆ POB
PO / AD = BO/ AB = 1/2
PO = AD/2 and CP = r - AD/2
Again
BO^2 - OP^2 =BP^2 =BC^2 -CP^2
BC^2 - r^2 = CP^2 - OP^2
= ( CP + OP) (CP - OP)
= r ( r - OP - OP)
= r ( r - AD)
Hereby 2 r ^2 - r AD = BC^2
Herein. 2 r ^2 - 7 r - 225 = 0
2 r^2 - 25 r + 18 r -225 = 0
(2 r - 25)( r + 9) = 0
Duameter of semi circle
2 r = 25
d=hypotenuse, make a line A to C, ACB will be 90 degrees. if CB = 15 than AC=20 and d=25. it takes ten seconds
Why?? Even if d actually is 25 you can‘t conclude from 15 to 20 to 25.
@@thomaslangbein2973-4-5 Triangle
You can use Ptolemy's theorem, and then get (d-25)(d+18)=0. So d=25. That's all.
I got as far as arcsin(3.5/r)x2+arcsin(7.5/r)x4=180.
I could only complete by trial and error as I had no idea how to isolate r. I was hoping to find out that there was a way to do that.
Very easy if cosine law is applied along the centre. No need for such complicated solution.
3 angles are 34/73/73?
Express BD in terms of d, hence cosC in terms of d. Also cosA = 7/d. Finally cosA = -cosC.
DC = BC results in
angle DOC = angle BOC
Hereby ∆ DOE & ∆ BOE congruent
angle DEO = angle BEO = π/2
Sin alterar las premisas del problema, se puede reordenar el esquema inicial de forma que quede una figura simétrica con la cuerda de 7 unidades de longitud, horizontal y centrada en la parte alta del semicírculo 》El triángulo de la derecha (de base "r" y lados "r" y 15) se compone de dos triángulos rectángulos con el cateto vertical común, hipotenusas "r" y 15 y bases (7/2) y (r -7/2) 》 r^2 - (7/2)^2 = 15^2 - (r - 7/2)^2 》 r=25/2 》 d=2r =2(25/2) =25=d
Gracias y saludos.
Look for triplets and this gives d = 25 also ptoelmy's theorem is satisfied with this...
Hey I was just wondering did you make use of one of the circle theorems? I ask this bc that might be the proof of why OD and OC are radii. I could be wrong.
Si fantástico uso de congruencias para llegar al buen resultado.
What this video lacks is a description of a plan to find the solution.
Νομίζω ότι δώσατε την πιο κατάλληλη λύση στο πρόβλημα αυτό. Η δική μου 1η σκέψη ήταν το θεώρημα του Πτολεμαίου (ACxBD=ADxBC+ABxCD) αλλά η αλγεβρική της επίλυση είναι δύσκολη από πλευράς πράξεων.
I did this reasoning, but the result is a little different. If the three chords of the semicircle were of the same length, it would be half the size of a regular hexagon. So I divided the overall length (15+15+7)/3=12.333...
In a regular hexagon inscribed in a circle, the side is equal to the radius. What's wrong?
Please make a different playlist for junior math Olympiad
You made simple into complicate.
Too algebraic, can solve it more geometrically.
Extend ad and bc, meet at e, you got 2 similar triangles, note the mid line, rest is pce of cake.......
Is that any different solution?
I do not have a good English.
Let be x and y the diagonals of the cyclic quadrilateral. So:
d^2-49=x^2
d^2-225=y^2.
But as the quadrilateral is cyclic xy=15*(7+d) ...(xy)^2=15^2*(d+7)^2
(d^2-49)*(d^2-225)=15^2*(d+7)^2
(d-7)*(d^2-225)=225*(d+7)
d^3-7d^2-450d=0 as d0
d^2-7d-450=0 d=25 or d=-18(not good)
So d=25.
Very nice solution
You are making this way harder than it needs to be. Just stop.
I passed through obscene amounts of trigonometry ( 15 = 2 r sin α/2 , 7 = 2 r sin (90° - α) and so on) but anyway I came through with the right solution😊 (very inelegant,but it worked).
Elegant solution!
Fantastic.
Solution without angles. Use "diagonals product" (e*f) theorem for cyclic quadrilaterals:
e*f = a*c + b*d
e*f = 15*2r + 7*15
Combine with two right triangle equations having 2r as their hypoteneuse:
e^2 = (2r)^2 - 15^2
f^2 = (2r)^2 - 7^2
Using some algebra, a quartic equation in r was developed, and solved using an online solver. r = 12.5. If Math Olympiad doesn't allow computational tools, it's growing obsolete as a forum for advanced problem-solving.
Great
my answer is 25 .2 simply using cosine law. 3 angles along the dia. are 34/73/73 degree. anyone agrees? why it diff. with yours (25)?
i made a mistake by proportioning the angles using the cord length. it is not correct. the three angles (along the centre) are proportioning by sine threta/2 vs the cord length. cord length = 2 r sin threta/2. using this relationship we can calculate the 3 angles and the r and hence the dia. the r = 12.5, dia = 25. thanks.
Gostei da solução
Is this guy trying to prove FLT or what?
Fine
very nice...
Here is a simpler solution
2*Arc(7/2) + 2*Arc(10/2) + 2*Arc(10/2) = 180 deg
2*A + 2*B + 2*B = 180
A +2*B =90................... SinA = Cos(2B)................... (1)
SinB = 15/d
SinA = 7/d= Cos(2B)= 1-2*(SinB ^2)
1-2*(15/d)^2 = 7/d
(d-25)*(d+18)=0
nice
Merci
Time passing video.
ab=2*(-225) 😎
Bubba... WHY DID YOU MAKE ME WATCH THIS
Taking the appropriate right triangle:
D² = (2.R)² = 7² + C²
C² = 4R² - 7²
C²/4 = R² - 7²/4
(C/2)² = R² - 3,5²
Taking the other appropriate right triangle:
(C/2)² + (R-3,5)² = 15²
(C/2)² = 15² - (R-3,5)²
Equalling :
R² - 3,5² = 15² - (R-3,5)²
R² - 3,5² = 15² - ( R² - 7R + 3,5²)
R² - 3,5² = 15² - R² + 7R - 3,5²
2 R² - 7R - 15² = 0
R² - 3,5 R - 112 ,5 = 0
R = 12,5 cm
D = 25 cm. ( Solved √ )
Al ojo
Traza el segmento BD y
Traza el segmento AC
Entonces se forma 90° en D y en C
Por arco, segmentos iguales (15) arcos iguales.
=> digamos angA = 2w
Y ang B= w + €,
de tal manera que w apunta a 15 y € apunta a 7
Entonces en triangulo ADB
angA + € = 90°
2w +€ = 90°
Es decir por razones complementarias
Sen€ = Cos 2w
Además
Por triangulos rectángulos
Sen€ = 7/d
Senw= 15/d
De Sen€ = Cos 2w
7/d = 1- 2sen(w)^2
7/d = 1- 2(15/d)^2
Resolviendo
d^2 - 7d + 450 = 0
Very long process. Sorry
Assuming that the drawing is in scale, I measure length "d" with a school ruler, then I calculate the proportion according to the scale, and it gives me approximately 25 cm
d = 25 cm (Solved √ )
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