$\begingroup$

$ABC$ is a triangle where $AE$ and $EB$ are angle bisectors, $|EC| = 5$, $|DE| = 3$, $|AB| = 9$. Find the perimeter of the triangle $ABC$.

I realized that the length of the side $|DC| = 8$. In the $\triangle BEC$, we have special triangle $3-4-5$. This is where I'm stuck. Can I take your thinkings?

$\endgroup$ 14

3 Answers

$\begingroup$

$|DE|=3=u$, $|EC|=5=v$, $|AB|=9=c$.

The question is not that innocent, as it looks at first glance.

Yes, following the angle bisector theorem we can calculate that $|BC|+|AC|=15$ and $|AB|+|BC|+|AC|=24$, as it was shown in the other answers.

But the question is, is it possible to construct a triangle with claimed properties?

Let $r,R$ be the radii of inscribed and circumscribed circles of $\triangle ABC$, respectively, and let $\rho=\tfrac12(|AB|+|BC|+|AC|)=12$.

\begin{align} \triangle CEG:\quad r&=v\sin\tfrac\gamma2 ,\\ \text{on the other hand, in }\triangle ABC\quad r&=(\rho-c)\tan\tfrac\gamma2 ,\\ \text{hence }\quad \cos\tfrac\gamma2&= \frac{\rho-c}{v}=\frac35 ,\\ \sin\tfrac\gamma2&=\frac45 ,\\ r&=4 ,\\ \sin\gamma&=2\cdot\frac35\cdot\frac45=\frac{24}{25} ,\\ R&=\frac{c}{2\sin\gamma} =\frac{9}{2\cdot\tfrac{24}{25}} =\frac{75}{16} . \end{align}

So far, so good. But, the validity test: \begin{align} \frac{r}{R} &=\frac{4\cdot16}{75} =\frac{64}{75}\approx 0.85 \end{align}

shows that this ratio is bigger than the known maximum for triangles, $\max\frac{r}{R}=0.5$.

We can actually calculate the sided of this "triangle", given that the known $\rho,r$ and $R$ uniquely define the sides of the triangle as the roots of cubic polynomial

\begin{align} x^3-2\rho x^2+(\rho^2+r^2+4r\,R)x-4r\,\rho\,R&=0 ,\\ x^3-24x^2+235x-900 &=0 ,\\ x_1&=9 ,\\ x_{2,3}&= \tfrac{15}2\pm\tfrac{5\sqrt7}2\cdot i . \end{align}

Indeed, the sum of the two other "sides" is $15$, but they are not real.

Thus, the answer is: a triangle with claimed properties does not exist.

Edit

In the following illustration, $|AB|=9$, and the locus of points $C$ is the ellipse with foci $A,B$ and $|CA|+|CB|=15$. The green line is the locus of points $D\in AB:\ |CD|=8$, the red curve is the locus of points $E\in CD:\ |CE|=5,\,|ED|=3$, and the blue curve is the locus of the centers $I$ of inscribed circles of corresponding triangles $ABC$.

The blue and red curves does not intersect, and $E=I$ never holds.

$\endgroup$ $\begingroup$

A simpler proof that such a triangle cannot exist, as also claimed by @g.kov.

The locus of points $P$ such that $PC/PD=5/3$ is a circle of radius $7.5$ passing through $E$ and with center on line $CD$. Hence $A$ and $B$ must belong to that circle and $AB$ is a chord passing through $D$.

But the shortest such chord is the one perpendicular to $CD$ ($HI$ in diagram below), whose length is $12$. Hence $AB$ cannot have length $9$ as stated in the problem.

$\endgroup$ $\begingroup$

Well this is a pretty straightforward question.

Now, According to the angle bisector theorem:

Consider a triangle ABC. Let the angle bisector of angle A intersect side BC at a point D between B and C. The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC

We get $$AC/AD=BC/BD=5/3$$ $$AC=(5/3)AD$$ and $$BC=(5/3)BD$$ Note that $$AB=AD+DB=9$$

Hence $$AC+BC==(5/3)(AD=DB)=15$$

But circumference of Triangle $=AB+BC+CA=15+9=24$

$\endgroup$ 3

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

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