In Kiselev's Planimetry book, he asks the construction problem: In a given triangle, find a point from which its sides are seen at the same angle.

I'm somewhat sure that this involves using the construction of an arc that encloses a given angle through a segment (On a given segment AB, to construct a disk segment enclosing a given angle), which is proven in the text and which I've used for a majority of the other construction problems. I also know that there is no "given angle" here, but that it would have to be 120 degrees because, for the angles around the necessary point were all equal, it'd be 360/3 = 120, but I can't construct a 120 degree angle.

That's the information I have an I am stumped as to what to do, or how to even put it together. My intuition tells me that I won't be using the 120 degree fact at all, and instead will have to somehow construct three circles that all intersect at the necessary point, but I'm not certain what properties the circles should have.

  • $\begingroup$ (Note: Im the owner of this) This app on Mac (itunes.apple.com/us/app/geometry/id972453094?mt=12 + exercises here www.think-geometry.com if interested) is for making constructions and check angles etc, you can play on the constraints after the construction is made $\endgroup$ – Thomas Oct 9 '15 at 4:33

What's your problem with constructing a $120^\circ$ angle?

Anyway, you can find the Fermat point as follows: On each side, erect (to the exterior) an equilateral triangle. From the third vertices of these triangles, draw the circles through the other two vertices of the same equilateral triangle. These circles intersect in the desired point (provided no angle of the originally given triangle exceeds $120^\circ$).

  • $\begingroup$ My apologies; I misunderstood the directions from the instructor: assuming no angle in the triangle was greater than 120 degrees. I see your proof and completely understand it, including why the 3 angles would be congruent (because they are all inscribed angles in circles that subtend a chord that is also subtended by an angle in the equilateral triangle, thus all of these angles are congruent). The one thing I'm having trouble negotiating is why do the three circles intersect at a single point. I've done the construction and I can see that they do, but how can I mathematically prove it? $\endgroup$ – analysischallenged Oct 4 '15 at 21:20
  • $\begingroup$ Nevermind; I figured it out! Thank you though!! :) $\endgroup$ – analysischallenged Oct 4 '15 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.