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  1. An isosceles triangle is a single triangle where two of its sides have the same length. The single triangle formed by the two S2 lines and the dotted line would be an isosceles triangle (regardless of what the angle between S1 and S2 is -- S1 is not part of the isosceles triangle).
  2. First some background on geometry for those who have either never taken it or have forgotten it. Two triangles are congruent if they have the same shape and size. They may be reflected or rotated or translated relative to each other, but by applying these transformations it is possible to make the two triangles superimposable. Given that there are two triangles, it is well known that if one triangle has sides of length {S1, S2, S3} and the other also has sides of length {S1, S2, S3} then the two triangles are congruent (this is sometimes called the SSS or side-side-side theorem in geometry). Similarly, if one triangle has two sides of length {S1, S2} and the angle between those two sides is {A3}, then any other triangle that also has two sides of length {S1, S2} and an angle in between them of {A3} must be congruent with it (this is called the SAS or side-angle-side theorem). There is a similar ASA theorem if you know the length of one of the sides and the angles at each end of that side. In grade school, I was quite distraught that there was no theorem stating that if you knew that two of the sides and one of the angles that does NOT fall between the two sides are equal then the triangles are congruent (that's because it's not true!), which would naturally be called the A$$ theorem. So I made up a proof for such a theorem on my own. The proof is below, see if you can spot the flaw in it. Proof: Take two triangles where you know the lengths of sides S1 and S2 and you know the angle A1 that is opposite from side S1. Position the two triangles so that their sides with length S1 are overlapping, the sides of length S2 for each triangle are both emerging from the same end of the common S1, and the triangles are pointing in opposite directions, as shown in the attached figure. Now imagine a line passing from the A1 of one triangle to the A1 of the other triangle. Consider the triangle formed by this line and the two lines of length S2. We next take advantage of a theorem on isosceles triangles: if a triangle has two sides of the same length, then the two angles opposite of those sides must have the same angle. We therefore know that the two angles labeled A1a are the same. Since the two angles A1 are the same, and the two angles A1a are the same, then the remaining angles labeled A1b in the figure (which span from the imaginary line to the side of unknown length for each triangle) must also be the same: just A1 minus A1a. Considering the triangle formed by the imaginary line and the two sides of unknown length, we can again use a theorem on isosceles triangles: if two angles of a triangle have the same angle, then the sides opposite those angles must have the same length. That proves that the two sides that were originally of unknown length must actually have the same length. Then the two triangles must be congruent, by the SSS theorem.
  3. plasmid

    Here's my explanation of the answer
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