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Source: Indian national math olympiad(INMO)

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vs2662

#1 h Dec 18, 2017, 11:09 AM • 2 Y

Y by Adventure10, Mango247

Let ABC be a right-angled triangle with angle B = 90 degree. Let D be a point on AC
such that the inradii of the triangles ABD and CBD are equal. If this common
value is R and if r is the inradius of triangle ABC, prove that 1/R = 1/r +1/BD

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2016 posts

PROF65

#3 h Dec 19, 2017, 10:24 PM • 2 Y

Y by Adventure10, Mango247

lemma:
let $ABC$ a triangle with inradius r and $D$ a point of $BC$ ,if $ABD$ and $ACD$ have the same inradius r' then :
$1)$ the line connectig their incenters $I_1I_2$ cuts $AD$ at its midpoint .
$2)$ $\frac{1}{r'}=\frac{1}{r}+\frac{AD}{S}$ where $S$ is the area of $ABC$ .
$3)$ $\frac{I_1I_2}{BC}= \frac{r}{IE^*}$ where $E^*$ is the intersection of the perpendicular through the incenter of $ABC$ , $I$ and the semicircle with diameter $BC$ which doesn’t overlap the triangle
proof
I presume $1)$ is clear.
$2)$ $\frac{AB+BC+CA+2AD}{2}=\frac{AB+BD+DA}{2}+\frac{DA+AC+CD}{2} \implies \frac{S}{R}+AD=\frac{S_1}{r'}+\frac{S_2}{r'}=\frac{S}{r'} \implies \frac{1}{r'}=\frac{1}{r}+\frac{AD}{S}$
$3)$ Let $E$ the foot of $I$ on $BC$ we know that $EDI_1I_2$ is cyclic implies $\angle I_1EI_2 =\angle I_1DI_2 =90^{\circ}$ thus the homothety that sends $I_1$ to $A$ sends $E$ to $E^*$ whence $\frac{I_1I_2}{BC}= \frac{r}{IE^*}$
back to the problem
it suffices to prove that $BD^2=S$ or $2BD^2=BA.BC$ .
$\frac{\sin \angle D}{AB} =\frac{\sin \angle A}{BD}, \frac{\sin \angle D}{ BC} =\frac{\sin \angle C}{BD}\implies \frac{\sin ^2 \angle D }{AB.BC} = \frac{2\sin \angle A \sin \angle C }{2BD^2}$ so it remains to show that $2\sin \angle A.\sin \angle C =\sin^2 \angle D$ , $\frac{\sin \angle A}{BC}=\frac{\sin \angle C}{BA}=\frac{1}{AC}\implies 2\sin \angle A.\sin \angle C=\frac{BA.BC}{b^2}$ besides $\frac{R}{IE^*}=\frac{I_1I_2}{CA}=\frac{r-R}{R}\implies \frac{I_1I_2}{CA}=\frac{r}{EE^*} \implies \left(2\frac{r}{I_1I_2}\right )^2$ $=\left(\frac{2{EE^*}}{AC}\right)^2$ $=\frac{4EA.EC}{AC^2} =\frac{ (AB+AC-BC)(AC+BC-AB)}{AC^2} =\frac{2BC.AB}{CA^2}$ but $\sin\angle D= \frac{r}{\tfrac{1}{2}I_1I_2}$ hence $2\sin \angle A .\sin \angle C =\sin^2 \angle D$ therefore the result follows .
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This post has been edited 1 time. Last edited by PROF65, Dec 19, 2017, 11:24 PM
Reason: typo

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