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

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

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[MarieAnne.]

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*

[johnnath]

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.

[Ivan-Matematyk]

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.

[bpark]

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.

[shadrana]

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=
Consider triangle ACB,
d^2=AC^2+CB^2 Pythagoras
CA=
The product of the opposite sides;


Sum of the products of opposite sides
Product of diagonals =
Rule(1) means:-

square each side,
(15d+105)^2=(d^2-49)(d^2-225)


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.

[marioalb]

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 √ )

[mariopopesco]

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.

[piman]

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.

[holyshit]

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

[samsheerparambil]

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

[Grizzly]

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.

[jimlocke]

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. <BOG = <COG = <COF = <DOF. Also, ΔODE and OAE are congruent. <AOE = <DOE The sum of all 6 angles is 180°. Note that <AOE = arcsin(3.5/R) and <DOF = arcsin(7.5/R). So, or Replace the 90° by x and, by trying values of R, attempt to find the value of R which results in x being 90°. Let's start with R=10 and get x=117°. We know that we need to increase R to reduce the value of x, so try R=11 and x= 104.5°. Try R=12 and x=94.3°. Try R=13 and x=86°. We now know that R=13 is too high and R=12 is too low, so try R=12.5 and, as luck would have it, on my scientific calculator. So, R=12.5 and the diameter is 25.

[zdrastvutye]

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 yd<0 then 60
30
40 if yc<0 then 60
50 print dg
60 return
70 gosub 20:if yd<0 or yc<0 then else 90
80 r=r+sw:goto 70
90 dg1=dg:r1=r:r=r+sw:if r>sw*1000 then stop
100 r2=r:gosub 20:if dg1*dg>0 then 90
110 r=(r1+r2)/2:gosub 20:if dg1*dg>0 then r1=r else r2=r
120 if abs(dg)>1E-10 then 110
130 print "r=";r:p=sw:goto 150
140
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
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
300 gosub 240:xbn=xb:ybn=yb:gosub 270:xba=xbn:yba=ybn:next a
r=12.5

16% 12% 1.96% 6.72
>
run in bbc basic sdl and hit ctrl tab to copy from the results window

[lybcxds]

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

[deebd]

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.

[krishnamoyghosh]

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

[sarantiskalaitzis]

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.

[NPSpaceZZZ]

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.

[derwolf]

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

[sumithpeiris]

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