Wednesday, March 15, 2006

I See Double! Topic: Geometry. Level: AIME/Olympiad.

Problem: (Problem-Solving Through Problems) Triangles $ ABC $ and $ DEF $ are inscribed in the same circle. Prove that

$ \sin{A}+\sin{B}+\sin{C} = \sin{D}+\sin{E}+\sin{F} $

if and only if the perimeters of the given triangles are equal.

Solution: Let's make use of a common theorem, the Extended Law of Sines. This tells us that in a triangle $ XYZ $ circumscribed in a circle of radius $ R $, we have

$ \frac{\sin{X}}{x} = \frac{\sin{Y}}{y} = \frac{\sin{Z}}{z} = \frac{1}{2R} $

where $ x,y,z $ are the sides opposite angles $ X,Y,Z $ respectively. The proof of this is quite simple; try finding it.

So applying this to our triangles $ ABC $ and $ DEF $ with equal circumradii $ r $, we have

$ \sin{A} = \frac{a}{2r} $, $ \sin{B} = \frac{b}{2r} $, $ \sin{C} = \frac{c}{2r} $


$ \sin{D} = \frac{d}{2r} $, $ \sin{E} = \frac{e}{2r} $, $ \sin{F} = \frac{f}{2r} $,

where the sides are defined in the standard way. So

$ \sin{A}+\sin{B}+\sin{C} = \sin{D}+\sin{E}+\sin{F} $

if and only if

$ \frac{a}{2r}+\frac{b}{2r}+\frac{c}{2r} = \frac{d}{2r}+\frac{e}{2r}+\frac{f}{2r} $

$ a+b+c = d+e+f $,

which means their perimeters are equal, as desired. QED.


Comment: Equating perimeters is often a strange thing, because the relationship of equivalent perimeters isn't particularly useful. In this case, however, noticing that their circumradii are the same immediately leads one to think of the Extended Law of Sines, which works out nicely with the trigonometry in the condition.


Practice Problem: Prove the Extended Law of Sines.


  1. Wow, very nice problem and solution.

  2. In triangle ABC draw the circumcenter P. We have PA = PB = PC. Now consider the triangles PAB, PBC, PAC. Knowing that angle BPC is twice angle BAC, etc., we infer that triangle PAB is isosceles with angle APB = 2A, etc.

    Now, drop altitudes from P. We have three pairs of right triangles with angles A, B, C, and from then we derive the relations

    sin A = BC / (2R)
    sin B = AC / (2R)
    sin C = AB / (2R)

    Which rearranges to form the Extended Law of Sines. (This relies on a property of circles I cannot prove, unfortunately. =/ )