Philippines Maths Olympiad 2020 Problem | Geometry | Important Geometry Skills Explained 

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.

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.

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

Very easy if cosine law is applied along the centre. No need for such complicated solution.

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.

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 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 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.

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.

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

Love these geometrical problems

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

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

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.

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 √ )

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.

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

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.

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)

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

sin⁡(BDA) = sin⁡(φ) = 1 → AB = 25 → ∆ABD = pyth. triple (7-24-25)

Beautiful problem. Thank you Sir.

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

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 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?

You can use Ptolemy's theorem, and then get (d-25)(d+18)=0. So d=25. That's all.

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.

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.

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

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.

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 √ )

Si fantástico uso de congruencias para llegar al buen resultado.

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.

Is that any different solution?

Please make a different playlist for junior math Olympiad

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.

Look for triplets and this gives d = 25 also ptoelmy's theorem is satisfied with this...

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

DC = BC results in angle DOC = angle BOC Hereby ∆ DOE & ∆ BOE congruent angle DEO = angle BEO = π/2

Very nice solution

Elegant solution!

Great

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.

Fantastic.

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.

Fine

You made simple into complicate.

Merci

Gostei da solução

nice

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).

Teşekkürler.

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)?

Is this guy trying to prove FLT or what?

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.

ab=2*(-225) 😎

Time passing video.

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

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

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.......

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 √ )

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