Solving triangles

Geometry Level 3

True or false :

You can solve a triangle if you're given a side, its opposite angle and the product of the remaining two sides.

This problem is part of my set: Geometry

True False

2 solutions

A Former Brilliant Member
Feb 22, 2016

Using standard conventions, a, A and bc(=k^2) are known.
Hence, B+C and k/2R(=z) are known.
Also, the given data implies sinBsinC=z^2.
Note that 2sinAsinB=cosA+cos(B-C)<=1+cosA=2(cos(A/2))^2 that is, z<=cos(A/2)
That is, 2ksin(A/2)<=a.
If a<2ksin(A/2), the triangle can't be solved.
Otherwise, cos(B-C) is known and hence B-C is known (with an ambiguity if B and C are not equal). Hence, A, B, C and R are known (with a possible ambiguity).

You can even solve it this way.First note that using sine rule,R can be determined.Using R =(abc)/4 (area) , area of triangle can be found.Since we have the value of area and one side,the value of altitude corresponding to that side can be found using area=(1/2)(base)*(height).So now,having the value of R and altitude,we can use construction.Construct segment BC whose length is given.Construct perpendicular bisector of BC.With B as centre and radius R intersect perpendicular bisector at O which is circumcentre.Now construct circumcircle of triangle.Draw a line parallel to BC at a distance equal to altitude of triangle.Depending on number of points of intersections of the parallel line and circumcircle,we will get total number of possible triangle ABC.

Indraneel Mukhopadhyaya - 5 years, 3 months ago

Nice construction. Bonus: Can you think of a straight edge compass construction without manipulating the given data algebraically (like finding circumradius or area of triangle)?

A Former Brilliant Member - 5 years, 3 months ago

"Depending on number of points of intersections of the parallel line and circumcircle, we will get total number of possible triangle ABC." There may be two, one or non intersection. So two one or no solution. Thanks for your solution. I thought just a little more expansion for some one not familiar with geometry. I too gave the same solution in my report.

Niranjan Khanderia - 5 years, 3 months ago

Yes exactly,depending on the number of points of intersection,there may be one,two or no solution.

Indraneel Mukhopadhyaya - 5 years, 3 months ago

What does $k$ stand for in the solution?

milind prabhu - 5 years ago

$k^2= bc$

A Former Brilliant Member - 5 years ago

Prakhar Bindal
Oct 27, 2016

Use cosine rule for angle A We can get the sum of squares of remaining sides and product is also known. two equations and two unknows we can solve and get all 3 sides of the triangle

Solving triangles

2 solutions

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