Geometry problem(Again a Challenge!!)

In a triangle ABCABC,BAC=60,AB=2AC.\angle BAC = 60^{\circ},AB=2AC.Point P is inside the triangle such that PA=3,PB=5,PC=2.PA=\sqrt{3}, PB=5,PC=2. What is the area of triangle ABCABC?

This is for those who are missing geometry section.The solution which I know is extremely surprising and beautiful but I want an alternate solution because the idea of solution which I know is impossible for a normal student to strike in his mind, even might be tough for good students.I think it might be easy for brilliant users who are good at geometry.So I ask all of you for an alternate solution.I think there are solutions using trigonometry but they are very ugly involving big big equations and variables.Let's see who solve it first.!It might be annoying because I am starting many discussions nowadays,its because RMO is coming nearer.Thanks in advance!

Note by Kishan K
5 years, 11 months ago

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12 votes

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It's not annoying! I enjoy it, and welcome more :)

Calvin Lin Staff - 5 years, 11 months ago

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Geometry is not my best area, so thanks for posting these interesting problems!!

Michael Tong - 5 years, 11 months ago

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I have a hunch that the geometry section will be coming up soon.

A Former Brilliant Member - 5 years, 11 months ago

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The triangle is a 30-60-90 triangle since one angle is 60 which is give and one side is twice another side (precisely the side opposite to 30 must be half the hypotenuse). If we assign the value xx to the hypotenuse the side opposite to 30 must be x/2x/2 and that opposite to 60 must be 3.5x/23^.5*x/2. Then we can get an equation in xx using Heron's formula for the areas of the three smaller triangles the sum of which must be equal to the base * height of the 30-60-90 triangle. Solving this equation is a whole new ball game though . I doubt you could do it without Wolfram.

A Former Brilliant Member - 5 years, 11 months ago

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We know the triangle ABCABC is a 30-60-90 right triangle. We can use coordinates. Let a>0,C(0,0),A(a,0),B(0,3a),P(x,y).a>0, C(0,0), A(a,0), B(0, \sqrt{3}a), P(x,y). Using distance formula, we have 3 equations. It's easy to deduce that a414a2+37=0.a^4-14a^2+37=0. Hence a2=7±23.a^2=7 \pm 2\sqrt{3}. If a2=723.a^2=7 - 2\sqrt{3}., then 2 is the longest side of the triangle PACPAC. Hence PAC>60.\angle{PAC} > 60^{\circ}. Therefore the point P is outside of triangle ABCABC. So a2=7+23a^2=7 + 2\sqrt{3}, and the area is simply 32a2=732+3.\frac{\sqrt{3}}{2}a^2 = \frac{7\sqrt{3}}{2} + 3.. Notice that APC=120 \angle{APC} = 120^{\circ} by Law of Cosines, some creative mind may construct a pure geometric argument from this info.

George G - 5 years, 11 months ago

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Let X,Y,ZX, Y, Z be the reflections of point PP w.r.t the sides CA,AB,BCCA, AB, BC, then it's easy to show the triangle XYZXYZ is a 3-4-5 triangle. So it's a right triangle with ZXY=90\angle ZXY=90^{\circ}. Hence AC2=7+23AC^2=7+2\sqrt{3} since CX=2,XA=3CX=2, XA=\sqrt{3} and CXA=120.\angle CXA = 120^{\circ}.

Or, we can add areas of three triangles BYZBYZ (an equilateral with sides=5), XYZXYZ (3-4-5 right triangle), and XYAXYA (120 degree isoceles triangle with sides 3,3,3\sqrt{3},\sqrt{3},3 ) together which is two times the area of the triangle ABCABC.

George G - 5 years, 11 months ago

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Yeah..this was the basic idea behind the problem.One needs to reflect the point P about all the sides and then there will be a pentagon(not hexagon) forming with these vertices.Find the area of the pentagon in 2 different ways and you will get your answer without trigonometry or coordinate geometry. By this method,the solution will be short and very beautiful because it involves extreme creativity.!!

Kishan k - 5 years, 11 months ago

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@Kishan K By using spiral similarity can give the same result :)

Xuming Liang - 5 years, 10 months ago

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see that if we apply cosine rule - then we have a=3ba=\sqrt{3}b then ΔABC\Delta ABC is a right angle triangle with sides b;a and 2b so ACB=90\angle ACB=90 WE NAME ACP=x\angle ACP=x so BCP=90x\angle BCP=90-x we again use cosine rule- cosx=1+b24bcosx=\frac{1+b^{2}}{4b} and cos(90x)=a2214acos(90-x)=\frac{a^{2}-21}{4a} or sinx=a2214asinx=\frac{a^{2}-21}{4a} now (sinx)2+(cosx)2=1(sinx)^{2}+(cosx)^{2}=1 or (a221)216a2+(1+b2)216b2=1\frac{(a^{2}-21)^{2}}{16a^{2}}+\frac{(1+b^{2})^2}{16b^{2}}=1 simflifying it will get b414b2+37=0b^{4}-14b^{2}+37=0 putting a=3ba=\sqrt{3}b from this equation we will get the value of b now the area of the triangle X=12bcsin60=32b2X=\frac{1}{2}bcsin60=\frac{\sqrt{3}}{2}b^{2} so putting the values of b we will get the area of ABC. b=7+23b=7+2\sqrt{3} by sridhar acharya's theorem

krishan Chakraborty - 5 years, 11 months ago

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then is a right angle triangle with sides b;a and 2b so WE NAME so we again use cosine rule- and or now or simflifying it will get putting from this equation we will get the value of b now the area of the triangle so putting the values of b we will get the area of ABC. by sridhar acharya's theorem

Kuldeep Singh - 5 years, 11 months ago

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