Poland Math Olympiad Geometry Problem | 2 Different Methods
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/ @mathbooster
∠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 √ )
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
You have eyes which are very straightforward to get facts! Amazing.
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
*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.*
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.
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
good solution. I also trird in this way but fail. How many time you take in solving this
@@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
Actually I am a math olympiad aspirant and want to increase my problem solving strategy. Your problem solving strategy is very good.
@@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!
@@soli9mana-soli4953 thanks a lot
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.
I did the same.
@@ludmilaivanova1603 Tribute of a French russophile ! These Amerloques have no style !
Triangle BPC is isosceles, hence
Indeed, exactly as I did it. Congratulations.
@RAG981 Good luck.
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
△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
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
*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.*
Nice solution
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
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.
Eres muy bueno explicando.
Yo, que no entiendo inglés, te entiendo. 👍
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.
AP=x=5
∵二等辺三角形の底角は等しい.
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.
!!! 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.
(5)^2=25 {30°B+50°C+50°A}={130°BCA+50°D}=180°BCAD 50^130 50^60^70 5^6^7 5^6^3^4 5^3^3^1^4 2^3^1^1^2^2 1^3^2^1 3^2 (BCAD ➖ 4BCAD+2). {160°A+10°B+10°C}=180°ABC 10^10^160 5^5^5^5^80^80 2^3^2^3^2^3^2^3^2^3^2^3 1^1^1^1^1^1^1^3^1^1^1^1^1^2 1^3^1^2 32(ABC ➖ 3ABC+2).
5(AB)sin(100) = 2(AB)(BM)sin(30), where (AB) = length of line AB, M is the point of middle of BC. So, 5sin(100) = b, b = (BM).
cos(10) = b/x. x = 5sin(100)/cos(10). You know, sin(100) = sin(90 + 10). So, x = 5. Math Booster, how about my solution? Please, tell me!
5
X = 4.996
È troppo semplice
@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
Thank God!! You are very slow!! तुम बहुत धीमे हो