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0|0|0|||||An interesting isosceles triangle.|sarathian||12:49:09|10/06/2007|    Here is a beautiful puzzle from geometry for all members.%0D%0A     ABC is an isosceles triangle with AB %3D AC. A perpendicular PQ is drawn on AB at its  mid-point P  %28i.e. AP %3DPB%29%2C and this perpendicular line cuts AC at Q. It turns out  that QC %3D BC. What is the  the vertex angle %2F_BAC%3F The solution is to be obtained analytically  and exactly. No trigonometrical tables..No numerical solutions.  I assure you it is analytically solvable.
2|1|0|||||RE%3A An interesting isosceles triangle.|mr_homm||06:47:08|10/09/2007|Yes%2C this is a beautiful puzzle.  Here is my solution%3A%0D%0A%0D%0AExtend BC and PQ to meet at D.  Draw AD to complete triangle ADB.  Since PD is the perpendicular bisector of AB%2C ABD is isoceles%2C with AD %3D BD.  Since Q is on the bisector PD%2C it follows by symmetry that BQ %3D AQ%2C so that BQA is isoceles.  %0D%0A%0D%0ANow let angle BAQ %3D ABQ %3D a and angle CBQ %3D CQB %3D b %28because BC %3D CQ%2C so that BCQ is isoceles%29.  Then at angle AQB %3D pi - 2a%2C and this angle is supplementary with b.  Therefore%2C pi - 2a %2B b %3D pi%2C so that b %3D 2a.  But since BQ divides angle ABC into a and b%2C and since ABC is isoceles%2C the sum of the angles on ABC is a %2B 2%28a%2B2a%29 %3D 7a %3D pi.  Thus a %3D pi%2F7.  Perhaps there are other methods of solution as well.%0D%0A%0D%0AThank you for an interesting problem%21%0D%0A%0D%0A--Stuart Anderson
3|2|2|||||RE%3A An interesting isosceles triangle.|alexb||07:00:04|10/09/2007|%3EPerhaps there are other methods of solution as well. %0D%0A%0D%0AYes%2C of course.%0D%0A%0D%0ATriangle BCQ is isosceles. If angle BCQ %3D %26alpha%3B%2C %0D%0A%0D%0Aangle CBQ %3D %28%26pi%3B - %26alpha%3B%29%2F2.%0D%0A%0D%0AAngle ABC %3D %26alpha%3B%2C hence angle ABQ %3D %26alpha%3B - CBQ %3D 3%26alpha%3B%2F2 - %26pi%3B%2F2.%0D%0A%0D%0ATriangle ABQ is isosceles because PQ is the perpendicualr bisector of AB. Hence angle BAC %3D ABQ %3D 3%26alpha%3B%2F2 - %26pi%3B%2F2.%0D%0A%0D%0AOn the other hand%2C angle BAC %3D %26pi%3B - 2%26alpha%3B.%0D%0A%0D%0AEquating the two expressions gives 7%26alpha%3B%2F2 %3D 3%26pi%3B%2F2 so that%0D%0A%0D%0A%26alpha%3B %3D %26pi%3B%2F7.
4|3|3|||||RE%3A An interesting isosceles triangle.|mr_homm||10:57:59|10/09/2007|I see from your proof that my construction of point D was unnecessary.  I noticed that ADB was an isoceles triangle and used that to show that AQB was isoceles.  However%2C as you point out%2C the reasoning I applied to ADB would apply directly to AQB just as well.  I think I prefer your method%2C which has less extraneous construction.%0D%0A%0D%0AHere is a related problem%3A%0D%0A%0D%0AAn isoceles triangle ABC with AB%3DAC is tiled by two smaller isoceles triangles.  What are the possible values for angle A%3F  I believe there are exactly 3 solutions%2C one of which is of course pi%2F7.  The stipulation that all triangles be isoceles makes the solutions more direct than in the original problem%2C but adds the complication of cases.%0D%0A%0D%0A--Stuart Anderson
5|4|4|||||RE%3A An interesting isosceles triangle.|alexb||11:02:22|10/09/2007|And even more generally%2C there%27s a problem of tiling an isosceles triangle with isosceles triangles%3A%0D%0A%0D%0Ahttp%3A%2F%2Fwww.cut-the-knot.org%2Ftriangle%2F80-80-20%2Findex.shtml%0D%0A%0D%0A%28The rightmost problem%29%0D%0A%0D%0A%28Without the condition that the bases of the tiles lie on the legs of the given triangle%2C any triangle can be cut into 6 isosceles triangles.%29