Poland Math Olympiad Geometry Problem | 2 Different Methods

[@marioalb9726]

∠BAC = 360°-160°-(50°+30°+2*10°) ∠BAC= 100° Inscribed angle theorem: ∠BAC is exacty half of ∠BPC = 360°-160°=200°, this means that we can draw a circumference with center on point P, where points A,B,C are points of this circumference. Then, AP=BP=PC=AC then x = 5 cm ( Solved √ )

[@marioalb9726]

Inscribed angle theorem: ∠BAC=100° half of ∠BPC=200° ∠ABC=30° half of ∠APC=60° ∠ACB=50° half of ∠APB=100° ∠APC=160°-100°=60° Everything matches !!! Triangle APC is an equilateral triangle, x= 5 cm

[@changryu8128]

You have eyes which are very straightforward to get facts! Amazing.

[@thierrygermain5182]

Here we go! Hello, friends ! Following SINE RULE in triangle ABC : 5 / sin30 = BC / sin100 = BC / cos10 Following TRIG in triangle PHC (H middle point of BC) : cos10 = HC / x ===> x = HC / cos10 = ( BC / cos10 ) / 2 = 5 / 2sin30 = 5.

[@ludmilaivanova1603]

I did the same.

[@thierrygermain5182]

@@ludmilaivanova1603 Tribute of a French russophile ! These Amerloques have no style !

[@oscarcastaneda5310]

We think Alike, my solution was along the lines of the first method : ) In the end I arrived at 4x²sin²(80°) = 100sin²(80°) from which x = 5.

[@soli9mana-soli4953]

There is a simple solution: extending BA untill intersects the perpendicular CD (D point of intersection) drawing in this way a right triangle whose angles are 30°,60°,90° and setting CD = a, it follows that BC = 2a. And being BPC isosceles we can say that BM = MC = a with M middelpoint of BC. Now we can see that triangles ADC and PHC are congruent because have the same angles 10°,80°90° and CD = CH = a then AC = PC = 5

[@FrpBypass-n7i]

good solution. I also trird in this way but fail. How many time you take in solving this

[@soli9mana-soli4953]

@@FrpBypass-n7i I can tell you that it was the third method I tried. In general, I take all the time I need, also because at my age I no longer have any school exams to take. And geometry is pure fun, for which I don’t have to answer to anyone than me

[@FrpBypass-n7i]

Actually I am a math olympiad aspirant and want to increase my problem solving strategy. Your problem solving strategy is very good.

[@soli9mana-soli4953]

@@FrpBypass-n7i Thank you! The advice I can give you is to learn how to draw lines. A well-drawn line turns a problem into a triviality. Good luck with your Olympiads!

[@FrpBypass-n7i]

@@soli9mana-soli4953 thanks a lot

[@JoanRosSendra]

Si dividimos el ángulo BAC en 40-60 la línea divisoria pasará por el punto P porque el ángulo PBA = BAP = 40. Es un triángulo isósceles, por tanto AP = BP = X En el triángulo APC tenemos los ángulos PAC, APC y CAP iguales y con valor 60º. Por tanto, se trata de un triángulo equilátero y sus tres lados son iguales: X=X=5 X=5 Saludos

[@ناصريناصر-س4ب]

Triangle BPC is isosceles, hence

[@RAG981]

Indeed, exactly as I did it. Congratulations.

[@ناصريناصر-س4ب]

@RAG981 Good luck.

[@kateknowles8055]

Constructing an equilateral triangle QPC with QP = QC = PC = X Marking S where QP crosses AB , QA = X - 5 the angles in triangle SAQ will be S = 40 , A= 80 and Q = 60 degrees. In isosceles triangle SPB S = 40 , P = 100 and B = 40 degrees PB = X so PS= X also SQ = zero so X = 5 That was a nice surprise. Worth the effort. Thank you , Math Booster

[@うっちゃん-e8e]

△PBCにおいてBCをYとおくと正弦定理よりY/sin160°₌Y/sin20°₌Y/2sin10°cos10°₌X/sin10° よってX₌Y/2cos10°　△ABCにおいて正弦定理よりY/sin100°₌Y/cos10°₌5/sin30°₌5/(1/2)₌10 X₌1/2×Y/cos10°₌1/2×10₌5

[@imetroangola17]

*Solução 2:* Seja H o pé da perpendicular de BC em relação ao vértice P do ∆PBC. Logo, BH = HC, além disso, cos (∠PBC) = HC/PC cos 10° = HC/x HC = x cos 10° BC = 2 HC BC = 2x cos 10°. Sendo A = 100°. Usando lei dos senos no ∆ABC, temos: BC/sen A = AC/sen 30° 2x cos 10°/sen 100° = 5/(1/2) 2x cos 10°/cos 10° = 10 2x = 10 *x = 5.*

[@GabrieleIris-is7bg]

Extend BA to D so that the new triangle BDC is rectangle in D and a 30°, 60°, 90° so CD=1/2BC. Now in triangle ACD, angle ACD=60°-50°=10°, ADC=90° for construction and CAD=80°. In triangle BCP, isosceles, if we call M the midpoint of BC we have triangle MPC congruent to ACD so X=5

[@ناصريناصر-س4ب]

Let H be the perpendicular projection of point A on BC. We have CH=5sin40, BH=AH*√3=5√3cos40. From this BC=5sin40+5√3cos40=10*sin(40+60)=10sin(100). From this x=BC/2cos10=(10sin 100)/(2sin 100)=5.

[@EddieDraaisma]

Consider a circle with center P and radius x so that B and C are on it. Looking from point A, the center angle of segment BC equals 200 deg (=360-160). Angle BAC = 100 deg = 200/2 deg, so point A is also on the circle. Angle ABC = 30 deg, then angle APC = 60 deg. Angle ACP = 60, so triangle ACP is equilateral. Then X = AP = AC = 5.

[@jimlocke9320]

We can use Math Booster's method to obtain the equation found at 5:00 and then rearrange it as: x = (((5)(sin(100°))(sin(10°)))/((sin(30°))(sin(160°))) and use a scientific calculator to find that x = 5 to within the precision of our calculator. We wish to prove that x = 5 is an exact answer. We note that AC = 5, CP = x = 5, and

[@imetroangola17]

*Solução:* Facilmente, encontramos A=100°. Note que A = 2 ∠ACB. Pela lei do seno no ∆ABC: BC/sen 100° = 5/sen 30° BC/sen 80° = 5/(1/2) = 10 BC = 10 sen 80°. Usando a lei dos cossenos no ∆PBC: BC² = 2x² (1 - cos 160°) 100 sen² 80° = 2x² (1 - cos 160°) Como cos 160° = 1 - 2sen² 80°, então 100 sen² 80° = 2x² (1 - 1+ 2sen² 80°) 100 sen² 80° = 4x² sen² 80° 25 = x² *x = 5.*

[@MataniMath]

Nice solution

[@bilgihancelebi1911]

160 derecenin olduğu köşe çemberin merkezi 100 derecenin olduğu köşe çemberin üzerindeki çevre açı olur. Bu köşeleri birleştirince sağ tarafta eşkenarüçgen oluşur ve x=5 çıkar.

[@СтасМ-ъ8б]

!!! Короткое решение! : Из точки P отложим отрезок PN = x на прямую AC. Возможны 2 случая: 1) CN < 5. Тогда ∠PNC= ∠PCN=60°. ( т.к.PN=PC=x ) ∠BPN=100°. И ∠NBP= ∠BNP=40° (т.к. BP=PN=x ) ! Противоречие: И ∠ABP=40 , и ∠NBP=40°. 2) CN>5 Аналогично получим противоречие- И ∠NBP=40°, и ∠ABP=40°. Тогда точки N и A совпадают. Тогда очевидно треугольник APC равносторонний. Т.е. x=5.

[@ina-j2p]

AP=x=5 ∵二等辺三角形の底角は等しい.

[@Eror7403]

Eres muy bueno explicando. Yo, que no entiendo inglés, te entiendo. 👍

[@nenetstree914]

5

[@giuseppemalaguti435]

È troppo semplice

[@matthieudutriaux]

@marioalb9726 : best method ! My method : X=((5*cos(50/180*Pi)+5*sin(50/180*Pi)/tan(30/180*Pi))/2)/sin((160/2)/180*Pi) X=5/2*(cos(50/180*Pi)+sin(50/180*Pi)/(1/sqrt(3)))/sin(80/180*Pi) X=5*(cos(50/180*Pi)*1/2+sin(50/180*Pi)*sqrt(3)/2)/cos(10/180*Pi) X=5*(cos(50/180*Pi)*cos(60/180*Pi)+sin(50/180*Pi)*sin(60/180*Pi))/cos(10/180*Pi) X=5*cos((50-60)/180*Pi)/cos(10/180*Pi) X=5*cos(10/180*Pi)/cos(10/180*Pi) X=5

[@marcelojabuti3619]

Thank God!! You are very slow!! तुम बहुत धीमे हो