An equilateral triangle ABC circumscribes the circle with equation $x^2 + y^2 = a^2$. The side BC of the triangle has equation $x = -a$.

a) Find the equations of AB and AC.

b) Find the equation of the circle circumscribing ABC.

Note: Pls show working when you answer and try to explain how you obtain the answer. By the way I am just a Year 10 student who is stuck on a textbook question and doing self-study before school starts.

My thoughts: I don't really know where to start to be honest. I tried to draw a diagram. $x = - a$ is a vertical line (parallel to y-axis) and is tangent to the given circle whose center is at the origin and radius is $a$. After drawing the diagram, I drew a triangle so that it circumscribes the circle and it looks like I need trigonometry to find AC and AB. Is that right? Since there is a right angle at the x-axis, do I use cos for AC and BC?

What I tried to do just now: Looking at the diagram, I attempted to use trigonometry for finding the equation of AC first. Since the inner circle's centre is on the origin, divided by the x-axis, I named M as the midpoint of BC. BM = CM has length $(\frac{- \alpha}{2})$. Also since the original equilateral triangle is now halved, it is a right angled triangle (there is a right angle and the other two angles has to be half the right angle). Now I use cos(45) $(\frac{- \alpha \div2}{h})$ but i kinda failed :( [h, for AB]

  • $\begingroup$ Looks like someone asked the same question as me a few minutes ago hehe. Lol just realised $\endgroup$ – DragonReborn Jan 10 '16 at 4:15
  • 1
    $\begingroup$ Same comment applies: show us what you've done so far, so that we can better help you. No effort on your part probably means no effort on ours, either. $\endgroup$ – John Hughes Jan 10 '16 at 4:18
  • $\begingroup$ I'm stuck. I've explained what I did in the 'My thoughts' section. I need some guidance. I don't even know where to go next after drawing the diagram. $\endgroup$ – DragonReborn Jan 10 '16 at 4:20
  • $\begingroup$ The triangle is above x=-a. The top vertex, A is on y=? (x=0. Why?) Does this helps? $\endgroup$ – Moti Jan 10 '16 at 4:47
  • $\begingroup$ Why is that? Isn't x=-a a vertical line instead of horizontal? So shouldn't the triangle, with the top vertex, be pointing left lying on the x-axis (x,0)? The triangle sit on the base of the line tangent to the incircle. Hmm maybe I drew it wrong... $\endgroup$ – DragonReborn Jan 10 '16 at 5:03

We can first determine the side length of $ABC.$ Since equilateral triangle $ABC$ circumscribes the circle, they must have the same center - the origin. Let point $P$ be the x-intercept of the line $x = -a.$ Then $OP = a.$ We know that $OP$ is a third of the median of triangle $ABC$ since it is equilateral. So the median length is just $3a.$ Using the 30-60-90 ratios, the side length of the triangle must be $\sqrt{3}a.$

a) Let $B$ be above $C.$ The coordinates for $B$ are $(-a, \frac{\sqrt{3}}{2}a).$ The coordinates for $A$ are $(2a, 0)$ because $OA$ is twice $OP.$ The equation of $AB$ is $\boxed{y = -\frac{\sqrt{3}}{6}x + \frac{\sqrt{3}}{3}}.$ By symmetry, the equation of line $CB$ is $\boxed{y = \frac{3}{6}x - \frac{\sqrt{3}}{3}}.$

b) Notice that $OA$ is a radius of the circumcircle of triangle $ABC$. We already found that $OA$ is $2a.$ So the equation of the circumcircle of triangle $ABC$ is $\boxed{x^{2} + y^{2} = 4a^{2}}.$ Notice that its center is also at the origin.

  • $\begingroup$ Thx so much! It was very helpful :) $\endgroup$ – DragonReborn Jan 10 '16 at 6:16

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.