Canada Math Olympiad Problem | A Very Nice Geometry Challenge

Canada Math Olympiad Problem | A Very Nice Geometry Challenge
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[@hongningsuen1348]

An alternative method using 1 construction only:
1. Let r be the radius of the semicircle.
2. Let C on PQ be the centre of the semicircle hence PC = CQ = r.
3. Join C to S with CS perpendicular to OB (radius to tangent) and CS = r
4. Join O to Q with angle BOQ = 45 (as arc AQ = arc BQ)
5. Let intersection of CS with OQ be D, forming triangle OSD with 45-45-90 angles.
OS = 3 (given)
DS = 3 (equal sides of isosceles triangle)
OD = sqrt(2) x 3 (Pythagoras theorem)
6. Triangle PQO is similar to triangle CQD (AAA as CS//PO & common angle at Q).
Hence PQ/CQ = PO/CD = QO/QD (proportionality equations for corresponding sides)
2r/r = PO/(CS - DS) = QO/(QO - OD)
2 = PO/(r - 3) = QO/(QO - sqrt(2) x 3)
Hence PO = 2(r - 3) and QO = sqrt(2) x 6
7. By cosine law for triangle OPQ with angle POQ = 45:
PQ^2 = OP^2 + OQ^2 - (2)(OP)(OQ)cos POQ
(2r)^2 = [2(r - 3)]^2 + [sqrt(2) x 6]^2 - 2[2(r - 3)][sqrt(2)x6]cos 45
4r^2 = 4(r^2 - 6r + 9) + 72 - 24(r - 3)
48r = 180
r = 15/4
8. Area of semicircle = 1/2 x (15/4)^2 x pi = (225/32)pi

[@WahranRai]

In Europe and the rest of the world (except for you) the angles in geometry and trigonometry are expressed with Greek letters alpha, beta, gamma... and the radius of a circle by R or r.
Why not using this standard notation

[@alexbayan8302]

This is probably the easiest Olympiad problem in a long while. My solution is almost the same and done in less than 5 min. It is still a nice problem and a confidence booster.

[@MdAli-jc3fe]

From how many years you have been practicing?😴

[@Irtsak]

Let QE⊥OA and QK⊥OB (construction)
Q is the middle of arc AB => EQ=QK (1)
(a point on the bisector of an angle is equidistant from the sides of the angle)
Construct CS. CS⊥OB (tangent of the semicircle) and S is the middle of OK (notice that OPQK is trapezium and CS is it’s median. So ΟS=SK=3 => ΟK=ΟS+SK=3+3=6 =>
EQ=6 (OKQE is rectangle) => QK=6 cause (1)
PS/SQ=OS/QK=3/6=1/2 => PS=SQ/2 (2)
Pythagoras theorem in right triangle PQS => QS²=QK²+SK² =>
QS² = 6² + 3²
QS²=45
Pythagoras theorem in right triangle PQS => PQ²=SQ²+PS² =>
(2R)² = QS²+ QS²/4=>
4R²=(5•QS² )/4 =>
R²=(5•45)/16
R²=225/16
Area of semicircle = (πR² )/2=225π/32

[@joseluishablutzelaceijas928]

Thank you for the problem and the solution. An alternative solution would be: To exploit the fact that Q is the midpoint of the arc AB, one can draw T as shown in the video and note that, as C is the midpoint of PQ, |ST| = 3 and |OT| = 6 = |TQ|. This means by Pythagoras that |QS| = 3*sqrt(5). Furthermore, as triangle PSQ is similar to triangle POS (to see this note that [angle QPS] = [CSP] = [OPS], from which the equality of the three corresponding angles in each triangle follows), one has that |PS|:|PO| = 3*sqrt(5)/3 = sqrt(5), which by similarity also implies that |PQ| = 5*|PO|. Applying Pythagoras on triangle POS one then obtains that |PO|^2 = 9/4 and the area of the semicircle is equal to (pi*(|PQ|/2)^2)/2 = 25*pi/8*|PO|^2 = 225*pi/32.

[@michaeldoerr5810]

I actually think that the above explanation shown in this video is more intuitive. More intuitive than the first two comments. No offense.

[@santiagoarosam430]

OQ es radio a 45°; T es la proyección ortogonal de Q sobre OB; QT corta al semicírculo en V y PV es paralela a OB; PQ=2r y su proyección horizontal OT=OS+ST=3+3=6 =QT.
Potencia de T respecto al semicírculo=3*3=VT*QT=6*VT; VT=9/6=3/2; QV=6-(3/2)=9/2.
En el triángulo QVP: (9/2)^2+(6^2)=(2r)^2; r^2=225/16; Área del semicírculo =225π/32.
Gracias y saludos.

[@quigonkenny]

Let M be rhe center of the semicircle. As radii, MS = MP = MQ = r. Draw MS. As OB is tangent to semicircle M at S and MS is a radius, ∠OSM = 90°.
Drop a perpendicular from Q to T on OB. Draw PL, where L is the point on MS where PL is perpendicular to MS. Draw MN, where N is the point on QT where MN is perpendicular to QT. MS and QT are both perpendicular to OB, and Therefore parallel to each other. PL and MN are perpendicular to MS and QT and thus parallel to each other. That all means that ∠LMP = ∠NQM and ∠MPL = ∠QMN, and as MP = MQ = r, ∆PLM and ∆MNQ are congruent. Thus ST = OS = 3.
Draw radius OQ, length R. As arc AQ = arc QB, ∠AOQ = ∠QOB = 90°/2 = 45°. As ∠QOT = 45° and ∠OTQ = 90°, ∠TQO = 90°-45° = 45°, ∆OTQ is an isosceles right triangle, amd TQ = OT = 6.
In triangle ∆MNQ, MN = 3, MQ = r, and as TQ = 6 and TN = MS = r, NQ = 6-r.
Triangle ∆MNQ:
NQ² + MN² = MQ²
(6-r)² + 3² = r²
36 - 12r + r² + 9 = r²
12r = 36 + 9 = 45
r = 45/12 = 15/4
Semicircle M:
A = πr²/2 = π(15/4)²/2 = 225π/32 sq units