Point on hypotenuse making the two inradii equal

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Point on hypotenuse making the two inradii equal

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Source: INMO 2016 Problem 5

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#1 h Jan 17, 2016, 1:08 PM • 1 Y

Y by Adventure10

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
$\cfrac{1}{r'}=\cfrac{1}{r}+\cfrac{1}{BD}.$

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#2 h Jan 17, 2016, 4:23 PM • 11 Y

Y by Re1gnover, Ankoganit, TheOneYouWant, YESMAths, anantmudgal09, darthsid, cr1, Viswanath, enhanced, Adventure10, Mango247

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

$BD=BD^2 \cdot \left (\frac{1}{r'}-\frac{1}{r} \right) \Longrightarrow \ \frac{1}{r'}=\frac{1}{r}+\frac{1}{BD}.$

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#3 h Jan 17, 2016, 4:54 PM • 2 Y

Y by Adventure10, Mango247

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

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#4 h Jan 17, 2016, 11:29 PM • 4 Y

Y by anantmudgal09, TheOneYouWant, AnArtist, Adventure10

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

$\frac{1}{\varrho}-\frac{1}{r}=\frac{AD}{[ABC]}=\frac{AD}{AD^2} \cdot \cot \frac{A}{2}=\cot \frac{A}{2} \cdot \frac{1}{AD} \Longrightarrow \frac{1}{\varrho}=\frac{1}{r}+\cot \frac{A}{2} \cdot \frac{1}{AD}.$

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#5 h Jan 18, 2016, 12:07 AM • 6 Y

Y by aayush-srivastava, Wizard_32, JayantJha, Anshul_singh, Adventure10, Mango247

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)

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#6 h Jan 18, 2016, 7:14 AM • 2 Y

Y by Adventure10, Mango247

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?

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#8 h Jan 18, 2016, 9:37 AM • 1 Y

Y by Adventure10

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

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#9 h Jan 18, 2016, 11:38 AM • 4 Y

Y by TheOneYouWant, YESMAths, Adventure10, Mango247

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

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#10 h Jan 18, 2016, 4:45 PM • 2 Y

Y by Adventure10, Mango247

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.

This post has been edited 1 time. Last edited by aditya21, Jan 18, 2016, 4:46 PM
Reason: e

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#11 h Jan 19, 2016, 8:40 AM • 19 Y

Y by PRO2000, AdithyaBhaskar, TheOneYouWant, biomathematics, tarzanjunior, darthsid, Viswanath, anantmudgal09, Ashutoshmaths, Ankoganit, Vrangr, Wizard_32, Kayak, neel02, lilavati_2005, amar_04, BVKRB-, Adventure10, Mango247

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

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#12 h Jan 19, 2016, 8:46 AM • 2 Y

Y by Adventure10, Mango247

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)...

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#16 h Mar 17, 2016, 6:04 PM • 2 Y

Y by PRO2000, Adventure10

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

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#17 h Apr 5, 2017, 4:40 AM • 2 Y

Y by Adventure10, Mango247

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

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#18 h Dec 31, 2017, 9:39 AM • 1 Y

Y by Adventure10

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

This post has been edited 1 time. Last edited by Wizard_32, Dec 31, 2017, 9:39 AM

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#19 h Dec 31, 2017, 10:33 PM • 1 Y

Y by Adventure10

http://artofproblemsolving.com/community/c6h1561314p9554167

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#20 h Feb 24, 2018, 11:51 PM • 2 Y

Y by Adventure10, Mango247

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 $\left( 1 - \frac{2r_1}{h} \right) \cdot \left( 1 - \frac{2r_2}{h} \right)= \left( 1 - \frac{2r}{h} \right)$ .

This post has been edited 2 times. Last edited by WizardMath, Feb 25, 2018, 12:22 AM

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#21 h Dec 25, 2018, 9:31 AM • 6 Y

Y by AlastorMoody, NOLF, lilavati_2005, Adventure10, Mango247, TSM_Zeki_MuREN

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. $\square$

This post has been edited 2 times. Last edited by Wizard_32, Dec 25, 2018, 9:47 AM

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#22 h Feb 5, 2022, 1:11 PM

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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
$\frac{1}{r'}=\frac{1}{r}+\frac{1}{BD}$ $\blacksquare$

This post has been edited 5 times. Last edited by NTistrulove, Feb 27, 2022, 7:10 PM

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