Angle Measurements of Triangles inside SemicircleDate: 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/ |
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