2
$\begingroup$

Let $\Delta ABC$ be a triangle in the plane. Let $P_{1}, P_{2}, P_{3}$ be the intersection points of bisectors, medians and altitudes, respectively. We define the quantity: \begin{equation} Q(\Delta ABC)=\frac{\mathcal{A}(\Delta P{1}P_{2}P_{3})}{\mathcal{A}(\Delta ABC)} \end{equation}
where $\mathcal{A}$ is the area of a triangle.

Is it true to say that $\;$ $\sup \{Q(\Delta ABC)\mid \; \Delta ABC\;\text{varies among all triangles}\}<1$?(strictly)

If yes, what type of triangles assumes this supremum?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ After thinking about it a bit more, there is a fairly easy formula for it. Specifically, if the angles are $\alpha,\beta,\gamma$, then up to a sign the ratio of the areas is the determinant of the matrix $((1,1,1),(\sin \alpha,\sin\beta,\sin \gamma)(\tan \alpha,\tan\beta,\tan\gamma))$ divided by $3(\sin \alpha+\sin\beta+\sin\gamma)(\tan\alpha+\tan\beta+\tan \gamma)$. In particular, in the case $\beta=2\alpha\to 0$ one sees that the ratio in fact goes to infinity. $\endgroup$
    – Lev Borisov
    Commented Jul 11, 2014 at 23:34
  • $\begingroup$ @LevBorisov thank you for your more attention on my question Thanks a lot for your beautiful formula. i think about your formula. $\endgroup$
    – Ali Taghavi
    Commented Jul 12, 2014 at 0:04

2 Answers 2

Reset to default
3
$\begingroup$

I can say something for the triangles with acute angles. But the obtuse angles may be a bit of a problem, since the intersection of the altitudes will be outside of the triangle.

For the triangle with acute angles, the supremum is definitely less than $1$. In fact, it would be true for any $P_1$ and $P_3$, so long as $P_2$ is the intersection of the medians. Indeed, if you drop the condition that $P_1$ and $P_3$ are intersection of bisectors and altitudes, then the supremum is independent of the triangle. More specifically, since area is linear in $P_1$ the supremum would be achieved when $P_1$ is a vertex of $ABC$, and similar for $P_3$. Thus, the ratio of the areas is less than $\frac 13$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ @LeviBorisov Thank you very much. I think 2/3. I am curious what is the exact value of this supremum and how is the shape of triangles of this value. $\endgroup$
    – Ali Taghavi
    Commented Jul 8, 2014 at 12:59
  • 1
    $\begingroup$ You may want to calculate what happens for a triangle with angles $\alpha,2\alpha,\pi-3\alpha$ as $\alpha\to 0$. $\endgroup$
    – Lev Borisov
    Commented Jul 8, 2014 at 15:01
0
$\begingroup$

There is a systematic method to attack you problem in the general case of three arbitrary given triangle centres (I am using the concepts and notations of the "Encyclopedia of Triangle Centers" which can be consulted online). We consider three centre functions $f$, $g$ and $h $ which we can assume to be homogeneous and have cyclic sum $1$. Then for the special triangle $ABC$ with vertices $(0,0)$, $(1,0)$ and $(p,q)$ ( the $(p,q)$-method), we have centres $$X_f=(f(b,c,a)+pf(c,a,b),qf(c,a,b))$$ etc. (This for the general case-- in your one it is probably quicker to compute them directly).The area of the internal triangle is then $1/2 ×$ the cyclic sum of $X_g\wedge X_h$. This can be expressed in terms of $p$ and $q$ by setting $c=1$ and replacing the remaining side lengths by $((p-1)^2+q^2)^{1/2}$ and $(p^2+q^2)^{1/2}$. The required area quotient is then obtained by dividing this by $q/2$. This reduces your question to ones about functions of two variables and these can be tackled by individual case.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .