We use standard notation $BC=a,$ $CA=b,$ $AB=c$ and $s=\tfrac{1}{2}(a+b+c).$ Thus letting $I_1$ and $I_2$ be the incenters of $\triangle ABD$ and $\triangle CBD,$ we get: $[ABC]=[BI_1D]+[BI_2D]+[ABI_1]+[CBI_2]+[AI_1D]+[CI_2D]=$ $=r' \cdot BD+\tfrac{1}{2}r' \cdot (a+b+c)=r' \cdot (BD+s) \Longrightarrow$ $BD=\frac{[ABC]}{r'}-s=\frac{[ABC]}{r'}-\frac{[ABC]}{r}=[ABC] \cdot \left (\frac{1}{r'}-\frac{1}{r} \right) \ (\star).$ But from Cono Sur 1994 (P6), we have $[ABC]=BD^2.$ Thus combining with $(\star)$ gives $BD=BD^2 \cdot \left (\frac{1}{r'}-\frac{1}{r} \right) \Longrightarrow \ \frac{1}{r'}=\frac{1}{r}+\frac{1}{BD}.$

Thanks Luis Gonzalez, almost everyone were stuck with the Cono Sur problem lemma...

Generalization: $D$ is a point on the side $\overline{BC}$ of $\triangle ABC,$ such that the inradii of $\triangle ABD$ and $\triangle ACD$ are equal to $\varrho.$ If $r$ is the inradius of $\triangle ABC,$ then we have the relation: \[ \frac{1}{\varrho}=\frac{1}{r}+\cot \frac{A}{2} \cdot \frac{1}{AD}\] As before, we have $[ABC]=r \cdot s=\varrho \cdot (AD+s) \ (\star)$ $\Longrightarrow \tfrac{2\varrho}{h_a}=\tfrac{a}{s+AD}.$ But from the infamous inradii problem, we have $1-\tfrac{a}{s}=\left (1-\tfrac{2\varrho}{h_a} \right)^2 \Longrightarrow$ $\frac{s-a}{s}=\left (1-\frac{a}{s+AD} \right)^2 \Longrightarrow AD= \frac{a \sqrt{s}}{\sqrt{s}-\sqrt{s-a}}-s=\sqrt{s(s-a)}=\sqrt{[ABC] \cot \frac{A}{2}}.$ Combining this latter relation with $(\star)$ yields $\frac{1}{\varrho}-\frac{1}{r}=\frac{AD}{[ABC]}=\frac{AD}{AD^2} \cdot \cot \frac{A}{2}=\cot \frac{A}{2} \cdot \frac{1}{AD} \Longrightarrow \frac{1}{\varrho}=\frac{1}{r}+\cot \frac{A}{2} \cdot \frac{1}{AD}.$

Here is my solution using some length chasing sort of arguments. This is quite hard for an average INMO problem but nevertheless, it can be done in an hour at the most at the contest. Let $AB=1$ and $BC=x$. Let $D$ be the point on $AC$ and let $\frac{AD}{DC}=k$. Let $[ABD]=\Delta_1$ and $[CBD]=\Delta_2$ and let $2s_1=AB+BD+DA$ and $2s_2=CB+BD+DC$. Now, we know that $\frac{r_1}{r_2}=1=\frac{\Delta_1(2s_2)}{\Delta_2(2s_1)}=\frac{AD}{DC}\frac{x+BD+DC}{1+BD+DA}$ which is equivalent to \begin{align*} \frac{AD}{DC}=k=\frac{1+BD}{x+BD} \end{align*} Now, this give that $BD=\frac{kx-1}{1-k}$ However, by Stewart's theorem $AC.BD^2=CD.AB^2+AD.BC^2$ and so computing this gives that $k=\frac{x+3}{3x+1}$. and similarly we get that $BD=\frac{(x+1)}{2}$. Now, $r=\frac{x+1-\sqrt{x^2+1}}{2}$ and that $r'=\frac{\Delta_1}{s_1}$. Now, $[ABC]=\frac{x}{2}$ and so we have that $\Delta_1(1+k)=[ABC]$ and so $\Delta_1=\frac{x(x+3)}{8(x+1)}$. Also, we know that $AD(1+k)=AC=\sqrt{x^2+1}$ and so $2AD=\frac{(x+3)\sqrt{x^2+1}}{2(x+1)}$ and hence, we have that \begin{align*} r_1=r'=\frac{\frac{x(x+3)}{8(x+1)}}{\frac{1+\frac{x+1}{2}+AD}{2}}=\frac{2x(x+1)}{(x+1)^2+2x+(x+1)\sqrt{x^2+1}} \end{align*} Now, we have that $\frac{1}{r}+\frac{1}{BD}=\frac{2}{x+1}+\frac{x}{x+1-\sqrt{x^2+1}}=\frac{2x(x+1)}{(x+1)^2+2x+(x+1)\sqrt{x^2+1}}$. This completes the proof $\blacksquare$ (This was just a sketch of what the expressions come out to be and is done by hand. It's not really hard to do all the intermediate steps, just that one needs a bit of computational fortitude)

anantmudgal09 wrote: ... just that one needs a bit of computational fortitude Unfortunatley, I don't have it. But anyway, I mentioned Stewart's theorem, and wrote down $r' = \frac{\triangle _1}{s_1} = \frac{\triangle _2}{s_2} = \frac{\triangle_1 + \triangle_2}{s_1 + s_2} = \frac{\triangle}{s + BD}$ and thus it is enough to prove $\triangle = BD^2$ (as$ \frac{1}{r} = \frac{s}{\triangle}$). And I fooled around a bit more with this, to add up to about 2 pages in total. How much, according to you, should that fetch me?

anantmudgal09 wrote: AdithyaBhaskar wrote: anantmudgal09 wrote: ... just that one needs a bit of computational fortitude Unfortunatley, I don't have it. But anyway, I mentioned Stewart's theorem, and wrote down $r' = \frac{\triangle _1}{s_1} = \frac{\triangle _2}{s_2} = \frac{\triangle_1 + \triangle_2}{s_1 + s_2} = \frac{\triangle}{s + BD}$ and thus it is enough to prove $\triangle = BD^2$ (as$ \frac{1}{r} = \frac{s}{\triangle}$). And I fooled around a bit more with this, to add up to about 2 pages in total. How much, according to you, should that fetch me? I don't really think that saying: "To proving $X$ it shall suffice to proving $Y$" and then not proving $Y$ is worth anything reasonable, at least not at INMO and certainly not at TST, not sure for RMO. Hey just observed: The result that I 'had to prove' is nothing but Cono Sur 1994/6

I think i finally have a 'not so bash ' sol to this problem Denote $AD=x,CD=y,BD=z$ Indeed,it suffices to show $z^2=ac/2$ Drop perps from $D$ on $AB,BC$ and use pyth to get $$z^2=(\dfrac{yc}{b})^2+(\dfrac{xa}{b})^2$$,that is, $$z^2=\dfrac{y^2c^2+x^2a^2}{x^2+y^2+2xy}$$. Now compare areas of $ABD,BDC$ by using formula $\Delta=rs$ and ratios of areas of triangle with same height.We get $$\dfrac{x}{y}=\dfrac{z+c+x}{z+a+y}=\dfrac{z+c}{z+a}$$. So $$z^2=\dfrac{(yc-xa)^2}{(x-y)^2}=\dfrac{y^2c^2+x^2a^2-2xayc}{x^2+y^2-2xy}$$Now compare this two equations for $z^2$, we get by componendo(or whatever you call that}$$z^2=2xayc/4xy=ac/2$$And done

though this question is less bashy than INMO 2015 P5, still .... my solution sketch = we easily get by $[ABC]=rs$ where $r$ is inradii. we get $\frac{1}{r'}=\frac{s+BD}{[ABC]}$ where $s$ is semi-perimeter of triangle $ABC$ and hence $\frac{1}{r'}=\cfrac{1}{r}+\cfrac{1}{BD}$ is equivalent to prove that $[ABC]=BD^2$ now since triangles $ABD,BCD$ have same inradii. than again using $[ABC]=rs$ we get $\frac{AB+BD}{AD}=\frac{BD+BC}{CD}$ which can than be bashed to get $2BD^2=AB.AC=2[ABC]$ as desired. so we are done.

'To bash or not to bash,that is the question' describes the geos of 2016 Inmo

Exactly @ above. Infact, on the overall the paper was so bashy, in the first 2 hours i got only 0.5 problems () since i was thinking up creative ideas for the others. However, in the last 2 hours reality struck and thankfully i was able to solve enough problems to pass(hopefully)...

Amusingly, it seems that something related to this was given in An Indian Practice test

anantmudgal09 wrote: Here is my solution using some length chasing sort of arguments. This is quite hard for an average INMO problem but nevertheless, it can be done in an hour at the most at the contest. Let $AB=1$ and $BC=x$. Let $D$ be the point on $AC$ and let $\frac{AD}{DC}=k$. Let $[ABD]=\Delta_1$ and $[CBD]=\Delta_2$ and let $2s_1=AB+BD+DA$ and $2s_2=CB+BD+DC$. Now, we know that $\frac{r_1}{r_2}=1=\frac{\Delta_1(2s_2)}{\Delta_2(2s_1)}=\frac{AD}{DC}\frac{x+BD+DC}{1+BD+DA}$ which is equivalent to \begin{align*} \frac{AD}{DC}=k=\frac{1+BD}{x+BD} \end{align*} Now, this give that $BD=\frac{kx-1}{1-k}$ However, by Stewart's theorem $AC.BD^2=CD.AB^2+AD.BC^2$ and so computing this gives that $k=\frac{x+3}{3x+1}$. and similarly we get that $BD=\frac{(x+1)}{2}$. Now, $r=\frac{x+1-\sqrt{x^2+1}}{2}$ and that $r'=\frac{\Delta_1}{s_1}$. Now, $[ABC]=\frac{x}{2}$ and so we have that $\Delta_1(1+k)=[ABC]$ and so $\Delta_1=\frac{x(x+3)}{8(x+1)}$. Also, we know that $AD(1+k)=AC=\sqrt{x^2+1}$ and so $2AD=\frac{(x+3)\sqrt{x^2+1}}{2(x+1)}$ and hence, we have that \begin{align*} r_1=r'=\frac{\frac{x(x+3)}{8(x+1)}}{\frac{1+\frac{x+1}{2}+AD}{2}}=\frac{2x(x+1)}{(x+1)^2+2x+(x+1)\sqrt{x^2+1}} \end{align*} Now, we have that $\frac{1}{r}+\frac{1}{BD}=\frac{2}{x+1}+\frac{x}{x+1-\sqrt{x^2+1}}=\frac{2x(x+1)}{(x+1)^2+2x+(x+1)\sqrt{x^2+1}}$. This completes the proof $\blacksquare$ (This was just a sketch of what the expressions come out to be and is done by hand. It's not really hard to do all the intermediate steps, just that one needs a bit of computational fortitude) How can you assume AB =1

themathfreak wrote: How can you assume AB =1 Because we are dealing with ratios. So if it is greater than (or less than) 1, the triangle would be just a scaled version of the assumed lengths (since $x$ can be anything!), and the similarity ratio would cancel out in the end, so it doesn't matter. This assumption just eases our work

http://artofproblemsolving.com/community/c6h1561314p9554167

My solution (from two years ago): Claim 1 : Let $ABC$ be a triangle with inradius $r$, and the point $D$ on $BC$ such that inradii of $ABD, ACD$ have a common inradius $r'$. If $s$ is the semiperimeter of $ABC$ and $BC = a$, then $AD = \sqrt{s(s-a)}$ Proof : Suppose $AD = d$, and the semiperimeters of $ABD$ and $ACD$ are $s_1, s_2$. $AB=c, AC=b$. Then $s_1+s_2=s+d$. Similar triangles and sum of areas yield $\frac{s+d}{s}=\frac{s_1+s_2}{s}= \frac{r}{r'}=\frac{s-b}{s_1-d} = \frac{s-c}{s_2-d} = \frac{a}{s_1+s_2-2d} = \frac{a}{s-d}$, so the claim follows. $\blacksquare$ Claim 2 : Let the height from $A$ onto $BC$ be $h$. Then $\left( 1 - \frac{2r'}{h} \right)^2 = \left( 1 - \frac{2r}{h} \right)$. Proof : We use a Lemma : $\frac{2r}{h} = \tan \frac{B}{2} \tan \frac{C}{2}$. Proof of lemma : $\tan \frac{B}{2} \tan \frac{C}{2} = \sqrt{\frac{(s-c)(s-a)}{s(s-b)}} \cdot \sqrt{\frac{(s-b)(s-a)}{s(s-c)}} = \frac {s-a}{s}$. But this is the ratio of the lengths of the tangents from $A$ to the incircle and the excircle, so the ratio must equal the distance of $A$ from the "topmost" tangent of the incircle and $BC$, which equals $\frac{h-2r}{h}$. $\blacksquare$ Now applying this lemma to $ABD$, $ACD$, and multiplying the two relations together, along with an application of the lemma to $ABC$ yields Claim 2. Now using this claim, we get $1-\frac{2r'}{h} = \sqrt{1-\frac{2r}{h}}= \sqrt{\frac{s-a}{s}} = \frac{AD}{s}$. We come back to the main problem. Note that $BD = \sqrt{s(s-b)} = \sqrt{\frac{ac}{2}}$. $\frac{1}{r'} - \frac{1}{BD} = \frac{(2s+h)\cdot BD - s h}{BD \cdot h \cdot (s-BD)} = \frac{2(ac+ab+b^2+bc) - \sqrt{2ac} \cdot (a+b+c)}{ac(a+b+c-\sqrt{2ac})} = \frac{2(ac+ab+b^2+bc) - \sqrt{2ac} \cdot (a+b+c)}{a^2b + ab^2 + abc + a^2c + ac^2 + bc^2 - ab^2 - b^3 -2ac\sqrt{ac/2})} = \frac{2(ac+ab+b^2+bc) - \sqrt{2ac} \cdot (a+b+c)}{(ab+ac+c^2-b^2)(a+b) - ((a+c)^2-b^2)\sqrt{ac/2})} = \frac{2(ac+ab+b^2+bc) - \sqrt{2ac} \cdot (a+b+c)}{(a+c-b)(a+b)(c+b) -(a+c-b)(a+b+c)\sqrt{ac/2})} = \frac{2}{a+c-b} = \frac{1}{r}$, wherein we have used $a^2 + c^2 = b^2$ and $h = \frac{ac}{b}$ and $r = s-b$. $\blacksquare$ Remark : This is essentially a brute force solution, but has an additional highlight of being able to find $BD$ generally, and also all the other lengths that are fruitful in this solution. Claim 2 can also be generalized, and for unequal radii $r_1, r_2$ instead of $r'$, we have $\left( 1 - \frac{2r_1}{h} \right) \cdot \left( 1 - \frac{2r_2}{h} \right)= \left( 1 - \frac{2r}{h} \right)$.

A year later, I tried this again, and successfully solved (bashed) this. I guess I have improved as a basher YESMAths wrote: Let $ABC$ be a right-angle triangle with $\angle B=90^{\circ}$. Let $D$ be a point on $AC$ such that the inradii of the triangles $ABD$ and $CBD$ are equal. If this common value is $r^{\prime}$ and if $r$ is the inradius of triangle $ABC$, prove that \[ \cfrac{1}{r'}=\cfrac{1}{r}+\cfrac{1}{BD}. \] Let $AD=x,$ and so $DC=b-x.$ Also, $AB=c, BC=a, BD=m.$ Then the condition gives $$\frac{[ABD]}{[BDC]}=\frac{s_{ABD}}{s_{BDC}} \Leftrightarrow \frac{x}{b-x}=\frac{c+x+m}{a+b-x+m}$$This simplifies to \begin{align*} x=\frac{b(c+m)}{a+c+2m} \end{align*}Now, $$r'=\frac{[ABD]}{s_{ABD}}=\frac{[BDC]}{s_{BDC}}=\frac{[ABD]+[BDC]}{s_{ABD}+s_{BDC}}=\frac{ac}{a+b+c+2m}$$Thus, it suffices to show $$\frac{1}{r'}=\frac{1}{r}+\frac{1}{m} \Leftrightarrow \frac{a+b+c+2m}{ac}=\frac{a+b+c}{ac}+\frac{1}{m} \Leftrightarrow 2m^2=ac$$By Stewart's theorem \begin{align*} & a^2x+c^2(b-x)=bm^2+bx(b-x) \\ &\Leftrightarrow (a^2-c^2)x+c^2b=b^2x-bx^2+bm^2 \\ &\Leftrightarrow x(-2c^2+bx)=b(m^2-c^2) & (\text{as }a^2+c^2=b^2 )\\ \\ &\Leftrightarrow \frac{b(c+m)}{a+c+2m} \left( -2c^2+\frac{b^2(c+m)}{a+c+2m} \right)=b(m^2-c^2) \\ \\ &\Leftrightarrow -2c^2(a+c+2m)+b^2(c+m)=(m-c)(a+c+2m)^2 \end{align*}This is not much computation, and the end expression is really neat: $$2m^3+2m^2a-mca-ca^2=0 \Leftrightarrow (2m^2-ac)(m+a)=0$$Hence, $2m^2=ac,$ as desired. $\square$

We know that $r=\frac{\Delta}{s}$ where $\Delta$ denoting the area of triangle. So we can say that, \[r'=\frac{\Delta_{CBD}}{s_{CBD}}=\frac{\Delta_{ABD}}{s_{ABD}}=\frac{\Delta_{CBD}+\Delta_{ABD}=\Delta_{ABC}}{s_{CBD}+s_{ABD}}\]So we have \[\frac{1}{r'}=\frac{s_{ABC}+BD}{\Delta_{ABC}}=\frac{1}{r}+\frac{BD}{\Delta_{ABC}}\] Claim: $BD^2=\Delta_{ABC}= AB\cdot BC/2$ Took help from ALM_04 for this part We have, \[\frac{\Delta_{CBD}}{\Delta_{ABD}}=\frac{s_{CBD}}{s_{ABD}}=\frac{CD}{AD}\]Therefore, \[\frac{s_{CBD}}{CD}=\frac{s_{ABD}}{AD}\]\[\frac{BC+BD}{CD}=\frac{AB+BD}{AD}\]After some intense bashing we will get the result $\square$ Therefore, we will have \[\frac{1}{r'}=\frac{1}{r}+\frac{1}{BD}\]$\blacksquare$