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I have a problem. In my triangle one vertex is in the vertex of the parabola and two others are in parabola. This is a isosceles triangle and I know one angle in this triangle : 120 grades. The question is: angle 120 grades must be beside of the vertex of parabola?

Thanks a lot for answer, I am beginner in math.

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Which option (1) or (2) is possible?

Nelliusz Frącek
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  • Is it given that the equal sides are AB and AC?(If C is the third point in your triangle) – Nemo Jan 29 '15 at 14:51
  • I updated question. – Nelliusz Frącek Jan 29 '15 at 15:01
  • If the sum of the three angles in a triangle is 180° and your triangle is isosceles with the equal sides at the vertex, then what other options do you have besides concluding that the 120 ° must be located at the vertex? In other words, I wouldn't mind to see how option 1 would work. – imranfat Jan 29 '15 at 15:14
  • I upadeted question, sorry for my not very good description. I know that second option is good, but first option (1 - y) is possible in some conditions? – Nelliusz Frącek Jan 29 '15 at 15:23

2 Answers2

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If by 120 you mean $120^\circ $ then it must be on the vertex of parabola. Proof is easy. Just take any point A on parabola, which is not vertex (let vertex point be point V) of parabola, then draw line from this point to vertex, then draw line from this point, such that it will form angle with $120^\circ $. You will get point B, which is intersection of new line with other leg of parabola. The two segments AV and AB are not equal, easy to prove.

Tahir Imanov
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choose the coordinates so that the equation of the parabola is $y = x^2.$ let the triangle be $OAB$ with $O =(0,0), A = (a, a^2), B= (b, b^2)$ and $\angle AOB = 120^\circ$ by cosine rule, $$AB^2 = OA^2 + OB^2 + OA*OB $$ so that $$(a-b)^2 + (a^2 - b^2)^2 = a^2 + a^4 + b^2 + b^4 + \sqrt{(a^2 + a^4)(b^2+b^4)} $$

cleaning it up a bit we end up with $$3a^2b^2 + 8ab - a^2 - b^2 + 3 = 0$$

which simplifies to $$-2ab(1+ab) = \sqrt{(a^2 + a^4)(b^2+b^4)} \tag 1$$ from $(1)$ we can see that $a$ and $b$ cannot be of the same sign. using symmetry we can assume that $a > 0, b < 0.$

a special solution is $-b= a = \sqrt{1+\sqrt 2}$ to find the general solution we will square $(1)$ and get a quadratic equation for $b$ $$b^2(3a^2 - 1) + 8ab + (3-a^2) = 0 $$ and the solution is $$ b = \dfrac{-2a \pm \sqrt 3 |a^2 -1|}{3a^2 - 1} \text{ and } b < 0.$$

if you take the negative solution you find that a negative solution exist for $0 \le < \sqrt 3$ and for the positive solution you have $\dfrac{1}{\sqrt 3} < a.$

abel
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