USA Math Olympiad Geometry Problem | Find the radius of the semicircle

[@ناصريناصر-س4ب]

We have PQ*PS=AP² and from it PS=9 so QS=QC=5 and according to Pythagoras' theorem we find PC=√41, AC=√5 the quadrilateral APQC is cyclic and from it 4*√5+6*5=√41*AQ and from it we find AQ=(30+4√5)/√41 we assume

[@JeanMarcBONICI]

concise and efficient, well done

[@marioalb9726]

Tangent secant theorem: 4.(4+c)=6² --> c= 6²/4-4= 5 cm Right triangle PQC: q² = 4² + c² ---> q= √41cm Right triangle PAC: a² = q² - 6² ---> a= √5cm Angle APQ: β= atan(c/4)+atan(a/6) = 71,7795° Cosine rule: p²=4²+6²-2*4*6*cosβ --> p= 6,08207cm Similarity of right triangles: 2R/p = q/4 R = p.q/8 = 4,868cm (Solved √)

[@marioalb9726]

Tangent secant theorem: 4.(4+c)=6² --> c= 6²/4-4= 5 cm Pytagorean theorem, right triangle PAC: d²= q²-6²= (4²+c²)-6² ---> d=√5 cm Cyclic quadrilateral APQC: p.q= a.c+b.d= 6*5+4*√5= 30+4√5 Similarity of right triangles: 2R/p = q/4 --> R = p.q/8 R= (15+2√5)/4 = 4,868cm (Solved √)

[@mohsenjalily2181]

Well done🙏🙏🙏🙏 Nice solution Post more videos

[@kycchen5825]

Let r be the radius . PC=√41, P, A ,C, Q four points co-circle By Ptolemy's theorem : QA*PC= 6*5+4√5. Let ∠CPQ = α= ∠QAC. Then Cos α =4/√41 and 2r Cos α = QA ⇨ r= (30+4√5)/8

[@sarantis40kalaitzis48]

Elegant Solution! Bravo.👏💯

[@TheAlavini]

Nice solution. Congrats. Deep offering problems like that.

[@RealQinnMalloryu4]

(6)^2(4)^2={36+16}=52 180°ABC/52=3.24ABC 3.2^12 3.2^6 3.2^2^3 1.1^2^3 2^3 (ABC ➖ 3ABC+2).

[@Rudepropre]

I used similarity and pythagoras and simultaneous equations

[@kateknowles8055]

Thank you for this problem .... I have done all this solving and still not successful.... time for viewing and reading... Angle AQC = angle BQS because angle AQB = angle CQS = 90º Extending QC to meet the lower semicircle at D will give a chord AD equal to SB. and SD is a diameter equal to AB . With O as midpoint of AB, AO = QO = SO = BO = DO = the radius of the semicircle. (circle) Angle QCB = Angle QPA they are 180 - angle ACQ Now why do QS and QC look like radii of another semicircle with SP as diameter? because this diagram is not drawn to scale This would make angle PCS =90º and PC=CS = 4sqrt(2) What would AC be ? In triangle PCA by Pythagoras' theorem AC.AC = PC.PC - AP.AP = 32 - 36 so this is nonsense and QS is not 4. One other construction: The perpendicular bisector of QS passes through O, through M midpoint of QS and through T where PT can be another tangent to the semicircle. PT = 6 QM.QM = (OT- OM)(OT+OM) by intersection of chords or by Pythagoras' theorem in triangle OMQ and OT= OQ QM.QM = OQ.OQ - OM.OM another false trail ? Triangles CAE and PQE are similar with E being the place where PA and QC meet CE/PE = AE/QE = CA/PQ = 1/ k suppose PQ = 4 so CA = 4/k I keep choosing similar triangles which pair up uselessly. Perhaps 3/10 marks for seeing something though! Leaving E to use again and drawing in F where PS will meet AB when they are both extended, there are some really big members of this set of similar triangles now. Angle P = theta = angle QCB The angle at E = The angle at F And time to start the viewing and reading Time for remembering another theorem that I had forgotten since way back when..... so 6.6 is 4.(4+5) so QS =5 and CS =5 ....... the only way to get forward....

[@JaimeTrespalacios-g2l]

Aunque el procedimiento es muy claro la traducción es pésima se le agregan expresiones que no tienen nada que ver con la solución del problema

[@KeithAllen-pg8ep]

First prize for ugliest geometry problem of 2025! Title should be: "Never Ending Solution."

[@giuseppemalaguti435]

CQ=a...con vari teoremi del coseno risulta a=5...=>r=√5/2+15/4=4,868...mah

[@FaridDjouher]

Non seulement vous fait du plagiat , vous explique l'exercixe au feux.