Inequality in a right triangle

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Inequality in a right triangle

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Source: INMO 2007 Question 1

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#1 h Nov 2, 2009, 11:57 AM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC$ right-angled at $C$ , the median through $B$ bisects the angle between $BA$ and the bisector of $\angle B$ . Prove that
$\frac{5}{2} < \frac{AB}{BC} < 3$

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#2 h Dec 12, 2009, 5:15 PM • 2 Y

Y by Adventure10, Mango247

The solution is a=2.65896...b
So a/b=2.65896... that is between 3/2 and 3

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#3 h Dec 12, 2009, 5:28 PM • 2 Y

Y by Adventure10, Mango247

castigioni wrote:

The solution is a=2.65896...b
So a/b=2.65896... that is between 3/2 and 3

Please write your full proof...I would really like to see it...

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#4 h Dec 28, 2009, 3:26 AM • 2 Y

Y by Adventure10, Mango247

Hey can't anyone give a full proof of this...
The 2nd post is definitely done by calculator....That is NOT ALLOWED in out national olympiads....

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behdad.math.math

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behdad.math.math

#5 h Dec 29, 2009, 8:23 AM • 2 Y

Y by Adventure10, Mango247

Rijul saini wrote:

In a triangle $ABC$ right-angled at $C$ , the median through $B$ bisects the angle between $BA$ and the bisector of $\angle B$ . Prove that
$\frac {5}{2} < \frac {AB}{BC} < 3$

$\angle\frac {B}{4}$ $=$ $\alpha$ and $M$ is the midpoint of $AC$
We can compute that $\sin3\alpha$ $\cos4\alpha$ $=$ $\sin\alpha$ ,
We know that $\frac {c}{sin90}$ $=$ $\frac {a}{sin(90 - 4\alpha)}$ , Hence $\frac {c}{a}$ = $\frac {1}{\cos4\alpha}$ = $\frac {\sin3\alpha}{\sin\alpha}$ = $3 - (\sin\alpha)^2$ .
$\Longrightarrow$ $\frac {c}{a}$ $<$ $3$ .
$\cos2\alpha$ $=$ $\frac {(\frac {c}{a} - 1)}{2}$ $\Longrightarrow$ $\frac {a}{c}$ $=$ $\cos4\alpha$ $=$ $\frac {1}{2}$ $(\frac {c}{a} - 1)^2 - 1$ .

Consider $\frac {c}{a}$ $\leq$ $\frac {5}{2}$ and this is contradict with the initial consideration, namely, $\frac {c}{a}$ $>$ $\frac {5}{2}$

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tc1729

#6 h Apr 29, 2012, 8:13 PM • 1 Y

Y by Adventure10

Since $E$ is the midpoint of $AC$ , we have $AE=EC=\frac{b}{2}$ . Since $BD$ bisects $\angle ABC$ , we also have $CD=\frac{ab}{a+c}$ . Since $BE$ bisects $\angle ABD$ , we have $\frac{BD^2}{BA^2}=\frac{DE^2}{EA^2}.$ However we have $\begin{cases}BD^2=BC^2+CD^2=a^2\frac{a^2b^2}{(a+c)^2} \\ DE^2=\left(\frac{b}{2}-\frac{ab}{a+c}\right)^2\end{cases}$ Using these in the above expression and simplifying yields $a\left[(a+c)^2+b^2\right] = c^2(c-a)^2.$ Using $c^2=a^2+b^2$ and eliminating $b$ , we obtain $c^3-2ac^2-a^2c-2a^3=0.$ Substituting $t=\frac{c}{a}$ , we reduce this to a cubic: $t^3-2t^2-t-2=0.$ Now, consider the function $f(t)=t^3-2t^2-t-2$ for $t>0$ , since $c/a\in\mathbb{Z}^{+}$ . For $\, we see that $f(t)=t^2(t-20-t-2<0$ . We also observe that $f(t)=(t-2)(t^2-1)-4$ is strictly increasing on the interval $(2, \infty)$ . It is easy to compute $f(5/2)=-\frac{11}{8}<0$ and $f(3)=4>0$ . Hence there is a unique value of $t$ in the interval $(\frac{5}{2}, 3)$ such that $f(t)=0$ . Hence $\frac{5}{2}<\frac{c}{a}<3$ as desired. $\Box$

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#7 h Jan 17, 2015, 2:42 PM • 1 Y

Y by Adventure10

my solution

solution =

$\angle\frac {B}{4} = \alpha$ and $M$ is the midpoint of $AC$
using sine law in triangle $ABE$ and $BEC$ , we obtain
that $\sin3\alpha \cos4\alpha = \sin\alpha ,$

We know that $\frac {c}{sin90} = \frac {a}{sin(90 - 4\alpha)}$ , Hence
$\frac {c}{a} = \frac {1}{\cos4\alpha} = \frac {\sin3\alpha}{\sin\alpha} = 3 - (\sin\alpha)^2.$
trivially,
$\Longrightarrow \frac {c}{a} < 3.$

now , to prove that $(5/2) < (c/a) = 3- 4sin^2\alpha$
this is equivalent to $sin\alpha< (1/2^{3/2} )$
now we know
$sin\alpha < sin (22.5) < (3^{1/2} - 1)/(2^{3/2}) < 1/(2^{3/2} )$

and we are done

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