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Angle Measurements of Triangles inside Semicircle


Date: 11/26/98 at 19:54:30
From: Dave Braha
Subject: Triangles inside semicircles

My calculus teacher gave our class this question about three weeks 
ago, but I keep getting stuck. The question says:

If the area of a triangle inside a semicircle is equal to the area 
outside the triangle within the semicircle, then find the values of 
the acute angles in the triangle.

So far I have come up with two different solutions: a triangle with 
angles 64, 26 and 90 degrees; and also one with 38, 38, and 104 
degrees. This leads me to believe that there could be a whole bunch 
of different solutions. 

Any help you could give me would be appreciated.

Thanks,
Dave


Date: 11/27/98 at 15:08:41
From: Doctor Rob
Subject: Re: Triangles inside semicircles

Since the radius of the circle is immaterial, let's work with the 
unit circle. Furthermore, since the location of the triangle is 
immaterial, let's put the longest side along the straight side of the 
semicircle. Let its length be L. Now the area inside the semicircle 
is Pi/2 square units, and the area inside the triangle should be half 
this, or Pi/4. Then if the altitude of the triangle is h, we have 
Pi/4 = h*L/2, or

   h = Pi/(2*L)

Since the triangle is inside the semicircle (0 < L <= 2), we have
Pi/4 <= h < infinity. But this also implies that h <= 1, so that
Pi/4 <= h <= 1, and Pi/2 <= L <= 2. For any L in this range, any 
triangle with altitude h = Pi/(2*L) and base L which fits inside the 
semicircle will work, such as an isosceles triangle with base L, the
midpoint of which is the center of the circle.

Indeed, there are infinitely many solutions.

If you insist that the vertices of the triangle are the endpoints of 
the diameter and a third point on the semicircle, so that all three 
vertices are on the semicircle, then L = 2, h = Pi/4, and there are 
two such triangles. The third vertex is either of the points, where 
the line parallel to the diameter and Pi/4 units from it intersects 
the circle. The angles are then uniquely determined. The angles at 
each end of the diameter have tangents

   4/Pi + sqrt[16/(Pi^2-1)]  and  4/Pi - sqrt[16/(Pi^2-1)]

Those angles are roughly 64.12124 degrees and 25.87876 degrees, and 
the third angle is 90 degrees exactly; this would be the first 
solution you found.

If all you need is for the three vertices to be on or within the 
boundaries of the semicircle, there are even more solutions!

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Conic Sections/Circles
High School Geometry
High School Triangles and Other Polygons

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