1
$\begingroup$

In a triangle $ABC$, point $X$ is picked on $BC$ such that the sum of the areas of the circumcircles of $ABX$ and $ACX$ is minimised. Describe where $X$ would be located on $BC$, and prove that this choice of $X$ is optimal.

Edit: Original problem statement said maximised, changed to minimised.

I would assume the point chosen would be a special point: midpoint, foot of altitude/angle bisector, etc., but don't know where to begin on a problem like this.

$\endgroup$
  • $\begingroup$ should not $X$ be either $B$ or $C$ depending on which angle is smaller? $\endgroup$ – abel Dec 31 '14 at 3:32
  • $\begingroup$ I think so Abel, At least in the degenerate case it holds $\endgroup$ – Jorge Fernández Hidalgo Dec 31 '14 at 3:34
  • $\begingroup$ Sorry, I meant finding the point where the sum is minimised, not maximised. Fixed. $\endgroup$ – minimario Dec 31 '14 at 3:38
  • $\begingroup$ for the minimizing question you may want to consider the midpoint, at least it works in the degenerate case. $\endgroup$ – Jorge Fernández Hidalgo Dec 31 '14 at 3:38
  • $\begingroup$ The midpoint definitely does not look minimal: prntscr.com/5mvg17 (Midpoint) prntscr.com/5mvg6u $\endgroup$ – minimario Dec 31 '14 at 3:43
2
$\begingroup$

here is the reason why $X$ must be either $B$ or $C.$ let $X$ be any interior point on $BC$ the diameter of the circumcircle $ABX$ is ${AB \over \sin \angle AXB}$ in the same way the diameter of the circumcircle $ACX$ is ${AC \over \sin \angle AXC}$ and $\sin \angle AXB = \sin \angle AXC$ plus the sum of the areas of the circle is ${\pi \over 4}{AB^2 + AC^2 \over \sin^2 \angle AXC}$ the maximum is achieved when this angle is the smallest. this happens at one of the extreme points $B$ or $C$ and minimum is achieved when $X$ is the foot of the perpendicular from $A$ to $BC.$

$\endgroup$
  • $\begingroup$ I think you forgot to square the sine in the denominator in the formula of the area, also the question now asks to minimize instead of maximize. But your method can easily be changed to answer the new question. The point should be the foot of the height. $\endgroup$ – Jorge Fernández Hidalgo Dec 31 '14 at 3:54
  • 2
    $\begingroup$ @JorgeFernández, thanks for catching the error. $\endgroup$ – abel Dec 31 '14 at 3:56
  • $\begingroup$ Excellent solution, but I think you've made a typo when you wrote ${AB \over \sin \angle AXC}$. You meant ${AC \over \sin \angle AXC}$, right? Other than that, everything looks good to me :) $\endgroup$ – minimario Dec 31 '14 at 4:10
  • $\begingroup$ @JamesLi, i fixed it. i was cutting and pasting it. $\endgroup$ – abel Dec 31 '14 at 4:14
  • $\begingroup$ I think there is another typo - (0.5AB and 0.5AC should be used to calculate the areas instead of using AB and AC directly). Also, I have a query, see post below. $\endgroup$ – Mick Dec 31 '14 at 5:21
0
$\begingroup$

This is not an answer but a query instead. It is too large to be stated in the comment.

First of all, I totally agree with @abel’s deduction.

However, when I tried to play with it using Geogebra, I have the following finding:-

enter image description here

A = (30, 80), B = (0, 0), C = (120, 0), X (now renamed as D) = (30, 0).

The blue circle ABD is drawn with $e: (x – 15)^2 + (y – 40)^2 = 1825$ (machine calculated, same for others)

The green circle ACD is also drawn with $f: (x – 75)^2 + (y – 40)^2 = 3625$

When the variable point D coincides with B, only one circle (the red one circumscribing ABC) needs to be drawn. Its equation is $g: (x – 60)^2 + (y – 23.13)^2 = 4134.77$

Clearly, $4134.77 < 1825 + 3625$.

This means the area of the single circle is less than the suggested minimal sum.

Would like to know what went wrong.

$\endgroup$
  • 1
    $\begingroup$ you are missing one more circle when the variable point $X$ coincides with $B$. the circle you are missing in tangent to $BC$ and has $BC$ for a chord. $\endgroup$ – abel Dec 31 '14 at 16:49
  • $\begingroup$ @abel Right. I thought that circle was a degenerated one with area = 0. Besides, error still exists after fixing. $\endgroup$ – Mick Jan 1 '15 at 5:11

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.