Poland Math Olympiad | A Very Nice Geometry Problem

Poland Math Olympiad | A Very Nice Geometry Problem
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[@user-bk9mn8de5c]

1)Pythagoras theorem for PQM: (PM)^2 +(QM)^2 = (PQ)^2
2)Pythagoras theorem for OQM: (OM)^2 +(QM)^2 = (OQ)^2
We do not need to calculate QM value. Instead 1) - 2)
(PM)^2 - (OM)^2 = (PQ)^2 - (OQ)^2
(8 - R)^2 - R^2 = (8 + R)^2 - (16 - R)^2
(8 - 2R)8 = -(8 - 2R)28
2R = 8
R = 4

[@onethreefiveeye]

10 seconds of thinking without thousand equations 😮

[@olesgaiduk5159]

Let's extend line PQ to the intersection with the small circle. This is point L. If we connect points PT L O we get a trapezoid. We construct a new circle around this trapezoid. Lines OT and P L pass through the center Q . Then TQ * QO = LQ * QP. That is, TQ = R; QO = 16- R; PQ = 8+ R; Q L = R . We have R * (16- R) = (8+ R) * R . We get 8 = 2 R . That is, R = 4.

[@richardleveson6467]

Nicely done - I always enjoy your constructions but seldom manage to complete them by myself!

[@soumighosh5382]

Small circle's diameter is semicircle's radius,
16/2 =8 is circle's diameter, 8/2=4 is its radius

[@piyathilakamuthunayaka7291]

I solved that nice problem. Thanks your channel.

[@quigonkenny]

As O is the center of a quarter circle and OA and OB are both radii, OA = OB = 16. Let P be the midpoint of OA and Q be the center of the small circle. As OA is the diameter of the semicircle, P is the center of the semicircle, and PA = PO = 8.
Let C be the point on OA where QC is parallel to OB, and let D be the point on OB where QD is perpendicular to OB. Draw PQ. As P and Q are the centers of the semicircle and circle respectively, the point of tangency between them will be on PQ, so PQ = 8+R. As Q is the center of the circle, QD = R. As QD and OA are parallel, and PO is the radius of the semicircle, PC = 8-R. Let QC = x.
Triangle ∆PCQ:
PC² + QC² = PQ²
(8-R)² + x² = (8+R)²
64 - 16R + R² + x² = 64 + 16R + R²
x² = 32R ---- [1]
Let OT be the radius of quarter circle O where OT passes through Q. As O and Q are the centers of the quarter circle and circle respectively, O, Q, and their point of tangency T will be colinear, so QT = R and OQ = 16-R. As QC and OB are parallel, QC = OD = x.
Triangle ∆ODQ:
QD² + OD² = OQ²
R² + x² = (16-R)²
R³ + 32R = 256 - 32R + R²

[@krishnaramachandran7722]

This is exactly the same method followed in the video!

[@gelbkehlchen]

Solution:
R = radius of the circle you are looking for,
r = radius of the quarter circle = 16,
M = center of the semicircle,
N = center of the small circle,
B = lower contact point of the small circle.
2 times Pythagoras:
(1) MN² = [(r/2)-R] ²+OB²
(2) ON² = R²+OB²
(1)-(2) = (3) MN²-ON² = [(r/2)-R] ²-R² ⟹
(3a) (8+R)²-(16-R)² = (8-R)²-R² ⟹
(3b) 64+16R+R²-(256-32R+R²) = 64-16R+R²-R² ⟹
(3c) 64+16R+R²-256+32R-R² = 64-16R |-64+16R+256 ⟹
(3d) 64R = 256 |/64 ⟹
(3e) R = 4

[@nunoalexandre6408]

Love it!!!!!!!!!

[@hayat.oyundur]

Cok uzun bir çözüm olmuş.
Yine de emeğiniz için teşekkürler 🙏

[@padraiggluck2980]

Thanks!

[@MathBooster]

Thank you for supporting this channel 🙂

[@MrLucaspingret]

No Brasil isto não existe por total falta de conhecimento

[@skwest]

Late to the party... but... here goes:
1) First off, Labels: Center of semicircle => P. Center of small circle => Q. Intersection of semicircle with small circle => C. Intersection of small circle with quarter circle => D. Intersection of small circle with OB => E.
2) Assign the length 'a' to segment EB, therefore the length of segment OE is (16 - a).
3) Connect points O and D (which length is the radius of the quarter circle, 16), passing through Q (I won't cite the supporting theora on these constructions). Noting that the length of segment OQ is (16 - R)22
4) We now have right △OEQ, with legs OE (16 - a) and EQ (R) and hypotenuse OQ (16 - R). Therefore, by Pythagoras, we have (16 - a)² + R² = (16 - R)². This is Equation #1.
5) Next we want to connect points P and Q (i.e. the centers of the semicircle and small circle) forming segment PQ of length (8 + R).
6) Now we need to project point Q onto segment OA at point S, creating segment SQ, parallel to radius OB. We note that segment SQ is the same length as segment OE, or (16 - a).
7) We now have right △QSP, with legs QS (16 - a) and SP (8 - R) and hypotenuse PQ (8 + R). Therefore, by Pythagoras, we have (16 - a)² + (8 - R)² = (8 + R)². This is Equation #2.
8) So, now we have 2 equations in 2 unknowns: #1 - (16 - a)² + R² = (16 - R)² and #2 - (16 - a)² + (8 - R)² = (8 + R)². Subtracting equation #2 from equation #1 yields equation #3:
R² - (8 - R)² = (16 - R)² - (8 + R)²
so that the "(16 - a)²" terms drop out.
9) Simplifying equation #3, yields (R = 4), which is our answer.
Now to watch the video to see if I got it right.
Cheers!

[@skwest]

Whoo hoo! Got it.
Thanks for the challenge.

[@gn01697933]

32R + R^2 = (16-R)^2

[@entp329]

R=4 by pythagoras 2times

[@runeaanderaa6840]

8+r=16-r 2r=8 r=4

[@frankzhou8617]

You are creating a new formula but I think it’s works when 16 changed to any number, well done!

[@duyphongtran8702]

R>r

[@runeaanderaa6840]

@@duyphongtran8702 True

[@user-sm1zb1pj8i]

請你證明為什麼

[@krishnaramachandran7722]

This is what you need to prove.

[@sadatsayem3724]

16/2=8/2= 4 where is the problem?

[@skwest]

Hmm... I think that the "8/2" step requires a little explaining.

[@dlspark7965]

R = OB/4

[@user-pb9ez2kh4q]

虽然一句也没听懂，但还是看懂了😂

[@prossvay8744]

R=4

[@mathstricknguon2246]

4

[@vahityurttas2234]

Ben 15 saniyede çözdüm

[@user-hh4yx4sq5o]

ㅇ

[@himadrikhanra7463]

8/3...?