Inscribed circle (St. Petersburg, 2002)

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Source: Bulgarian MO 2007, Day 1, Problem 1

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#1 h May 16, 2007, 3:39 PM • 2 Y

Y by Adventure10 and 1 other user

The quadrilateral $ABCD$ , where $\angle BAD+\angle ADC>\pi$ , is inscribed a circle with centre $I$ . A line through $I$ intersects $AB$ and $CD$ in points $X$ and $Y$ respectively such that $IX=IY$ . Prove that $AX\cdot DY=BX\cdot CY$ .

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#2 h May 22, 2007, 1:35 AM • 2 Y

Y by Adventure10, Mango247

bilarev wrote:

In the quadrangle $ABCD$ ( $\angle BAD+\angle ADC>\pi$ ) is inscribed a circle with center $I.$ A line throught $I$ intersects $AB$ and $CD$ in points $X$ and $Y$ respectively such that $IX=IY.$ Prove that $AX.DY=BX.CY.$

bilarev, i think that your problem, in this form, is wrong!
Your problem implies that M, N and I are in the same line, where M and N are the points where the incircle touchs AB and CD! (The congruence of the triangles MXI and NYI)

EDIT: I'm Wrong! See my solution in my next post!

This post has been edited 1 time. Last edited by e.lopes, May 22, 2007, 6:12 PM

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#3 h May 22, 2007, 6:16 AM • 2 Y

Y by Adventure10, Mango247

e.lopes wrote:

bilarev, i think that your problem, in this form, is wrong!
Your problem implies that M, N and I are in the same line, where M and N are the points where the incircle touchs AB and CD! (The congruence of the triangles MXI and NYI)

You are wrong...think a little bit about the positin of the points...(M and N may lie on segments AX and DY).

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#4 h May 22, 2007, 4:23 PM • 2 Y

Y by Adventure10, Mango247

bilarev wrote:

e.lopes wrote:

bilarev, i think that your problem, in this form, is wrong!
Your problem implies that M, N and I are in the same line, where M and N are the points where the incircle touchs AB and CD! (The congruence of the triangles MXI and NYI)

You are wrong...think a little bit about the positin of the points...(M and N may lie on segments AX and DY).

Thank you for your correction. (i wasn't using that A+D > 180)
Here is my solution, with the points in your correct places!

_

Let M and N the points where the incircle touchs AB and DC!

We know that the triangles MXI and NYI are congruents! So, $<MIX = <NIY = 90-\frac{MIN}{2}= \frac{B+C}{2}$ .

Let $MX = NY = x$ .

First, we know that $x = \frac{r}{tan(\frac{B+C}{2})}$ .

See that we have to prove that $(AM+x)(DN+x) = (MB-x)(NC-x)$ .

This is the same that prove that $\frac{r}{tan(\frac{B+C}{2})}= \frac{MB.NC-AM.DN}{MB+MA+NC+ND}$ .

We easily see that $r(MB+MA+NC+ND) = [ABCD]$ .

So, we have to prove that $[ABCD] = (MB.NC-AM.DN)tg(\frac{B+C}{2})$ .

We easily see that $[ADI]+[BCI] = [ABI]+[CDI]$ . So, $[ABCD] = 2([ADI]+[BCI]) = (AI.DI+BI.CI)sen(\frac{B+C}{2})$ .

So, we have to prove that $BM.CN-DN.AM = (AI.DI+BI.CI)cos(\frac{B+C}{2})$ .

By Cosines Law in the triangles AID and BIC:

$AI.DIcos(\frac{B+C}{2})+BI.CIcos(\frac{B+C}{2}) = \frac{(AI^{2}+DI^{2}-AD^{2})-(BI^{2}+CI^{2}-BC^{2})}{2}$

$= \frac{(r^{2}+AM^{2}+r^{2}+DN^{2}-(AM+DN)^{2})-(r^{2}+BM^{2}+r^{2}+NC^{2}-(BM+NC)^{2})}{2}$

$= BM.CN-DN.AM$ !

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yetti

#5 h May 22, 2007, 4:38 PM • 2 Y

Y by Adventure10, Mango247

Let $\rho$ be radius of (I). Let P, Q be tangency points of AB, CD with (I). Let AB, CD meet at E, on the opposite side of DA than I (because $\angle D+\angle A > \pi$ ). EI bisects $\angle DEA \equiv \angle CEB,$ $XY \perp EI.$ Reflect E in XY into E', EXE'Y is a rhombus, E'X, E'Y are tangents of (I) at P', Q', and E'P' = E'Q' = EP = EQ. Let DA meet E'X at F and E'Y at G. (I) is the E-excircle of the $\triangle DEA,$ F-excircle of the $\triangle FXA,$ and G-excircle of the $\triangle DYG$ and these 3 triangles are similar, having equal angles. Let $r_{d},\ r_{e}= \rho,\ r_{a}$ be exradii, r inradius, $s = EP = EQ$ semiperimeter, and $S = \sqrt{r r_{d}r_{e}r_{a}}$ area of the $\triangle DEA$ (Heron's formula for S). Then $\frac{AX}{EA}= \frac{r_{e}}{r_{d}},$ $\frac{DY}{ED}= \frac{r_{e}}{r_{a}},$

$AX \cdot DY = EA \cdot ED \cdot \frac{r_{e}^{2}}{r_{d}r_{a}}= \frac{2S}{\sin E}\cdot \frac{r_{e}^{3}r}{S^{2}}= \frac{2 r_{e}^{3}}{s \sin E}= \frac{2 \rho^{3}}{EP \sin E}$

Similarly, $BX \cdot CY = \frac{2 \rho^{3}}{E'P' \sin E'}= \frac{2 \rho^{3}}{EP \sin E},$ hence $AX \cdot DY = BX \cdot CY.$

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#6 h May 24, 2007, 4:27 AM • 2 Y

Y by Adventure10, Mango247

Sorry, this problem is equivalently with a nice and easy old problem :

Quote:

An equivalent enunciation. Let $C(I)$ be the incircle of the triangle $ABC$ . The points $E\in AC$ and $F\in AB$ belong to the circle $w$ . Denote the points $X\in (AC)$ and $Y\in (AB)$ for which $I\in XY$ and $IX=IY$ . Prove that for any points $M\in (AE)$ and $N\in (AF)$ for which the line $MN$ is tangent to the circle $w$ , the product $XM\cdot YN$ is constant, more precisely, $XM\cdot YN=IX^{2}$ .

Proof. Denote $\{\begin{array}{c}m(\widehat{BNM})=2y\\\ m(\widehat{CMN}=2x\end{array}$ . Observe that $\boxed{x+y=90^{\circ}+\frac{A}{2}}$ and $\{\begin{array}{c}m(\widehat{XIM})=90^{\circ}+\frac{A}{2}-x\ ,\ m(\widehat{XIC})=\frac{B}{2}\\\ m(\widehat{YIN})=90^{\circ}+\frac{A}{2}-y\ ,\ m(\widehat{YIB})=\frac{C}{2}\end{array}$ . Therefore,

$\{\begin{array}{c}\frac{XM}{XC}=\frac{IM}{IC}\cdot\frac{\sin\widehat{XIM}}{\sin\widehat{XIC}}=\frac{\sin\frac{C}{2}}{\sin x}\cdot\frac{\cos(x-\frac{A}{2})}{\sin\frac{B}{2}}\\\\ \frac{YN}{YB}=\frac{IN}{IB}\cdot\frac{\sin\widehat{YIN}}{\sin\widehat{YIB}}=\frac{\sin\frac{B}{2}}{\sin y}\cdot\frac{\cos(y-\frac{A}{2})}{\sin\frac{C}{2}}\end{array}$ $\implies$ $\frac{XM}{XC}\cdot\frac{YN}{YB}=$ $\frac{\cos(x-\frac{A}{2})\cos(y-\frac{A}{2})}{\sin x\sin y}=$ $\frac{\cos (x-y)+\cos (x+y-A)}{\cos (x-y)-\cos (x+y)}=$

$\frac{\cos (x-y)+\cos(90^{\circ}-\frac{A}{2})}{\cos (x-y)-\cos(90^{\circ}+\frac{A}{2})}=$ $\frac{\cos (x-y)+\sin\frac{A}{2}}{\cos (x-y)+\sin\frac{A}{2}}=1$ $\implies$ $XM\cdot YN=XC\cdot YB$ (constant).

Observe that the constant $k\equiv XC\cdot YB=XA\cdot XE=IX^{2}=$ $(IA\cdot\tan\frac{A}{2})^{2}=$ $\frac{bc(p-a)}{p}\cdot\frac{(p-b)(p-c)}{p(p-a)}=\frac{bc(p-b)(p-c)}{p^{2}}$ .

Remarks.

$1\blacktriangleright$ The points $X$ , $Y$ belong to the circle which is tangent to the rays $[AB$ , $[AC$ and which is internal tangent to the circumcircle of $\triangle ABC$ .

$2\blacktriangleright XA=YA=\frac{bc}{p}$ . Memorize the following relations $\{\begin{array}{c}p(p-a)+(p-b)(p-c)=bc\\\ p(p-a)-(p-b)(p-c)=bc\cdot\cos A\end{array}$ , where $a+b+c=2p$ .

$3\blacktriangleright$ From the above identities you can prove very easily the following remarkable formulas : $\{\begin{array}{c}\sin\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{bc}}\\\\ \cos\frac{A}{2}=\sqrt{\frac{p(p-a)}{bc}}\\\\ \tan\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{p(p-a)}}\end{array}$

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#7 h Mar 17, 2008, 1:31 PM • 2 Y

Y by Adventure10, Mango247

It is easy to show by angle chasing (but I am too lazy to write it), that $\angle AIX = \angle IDY$ and $\angle BIX = \angle ICY.$

So, from the similar triangles $\bigtriangleup AIX$ and $\bigtriangleup IDY,$ we conclude that $\frac {AX}{IY} = \frac {IX}{DY}$ $\Longrightarrow$ $(AX)\cdot (DY) = (IX)^{2}$ $,(1)$

Also, from the similar triangles $\bigtriangleup BIX$ and $\bigtriangleup ICY,$ we conclude that $\frac {BX}{IY} = \frac {IX}{CY}$ $\Longrightarrow$ $(BX)\cdot (CY) = (IX)^{2}$ $,(2)$

Hence, from $(1),$ $(2)$ $\Longrightarrow$ $(AX)\cdot (DY) = (BX)\cdot (CY)$ and the proof is completed.

Kostas Vittas.

Attachments:

t=149151.pdf (4kb)

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#8 h Jul 19, 2010, 9:08 PM • 1 Y

Y by Adventure10

In case one is being too stupid to notice the similar triangles, this problem can also be bashed easily.

Let the lengths of the tangents to the incircle from vertices $A, B, C, D$ be $a, b, c, d$ , respectively. Let $AB,CD$ touch the incircle at $M,N$ , respectively.

Lemma: If $r$ is the inradius of tangential quadrilateral $ABCD$ then
$r^2 = \frac {abc + abd + acd + bcd}{a + b + c + d}.$
Proof: We can easily see that $A + B + C + D = 2\pi$ , $\tan(A/2) = r/a$ , $\tan(B/2) = r/b$ , $\tan(C/2) = r/c$ , and $\tan(D/2) = r/d$ . Then
$0 = \tan\pi = \tan\left(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}+\frac{D}{2}\right)\\ = \frac {\sum\tan\frac{A}{2} - \sum\tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}}{1 - \sum\tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}\tan\frac{D}{2}},$ so clearing the denominator,
$\frac{r}{a} + \frac{r}{b} + \frac{r}{c} + \frac{r}{d} = \frac{r^3}{abc} + \frac{r^3}{abd} + \frac{r^3}{acd} + \frac{r^3}{bcd}.$ and the lemma follows. $\blacksquare$

Now let $MX=NY=e$ . We need to show
$AX\cdot DY=BX\cdot CY \Longleftrightarrow (a+e)(d+e)=(b-e)(c-e) \Longleftrightarrow e=\frac{bc-ad}{a+b+c+d}.$ Note that
$\begin{align*} \frac{e}{r}=\tan\angle{MIX}=\tan\frac{A+D-\pi}{2}=-\frac{1}{\tan\frac{A+D}{2}}&=\frac{\tan\frac{A}{2}\tan\frac{D}{2}-1}{\tan\frac{A}{2}+\tan\frac{D}{2}}\\ &=\frac{\frac{r}{a}\cdot\frac{r}{d}-1}{\frac{r}{a}+\frac{r}{d}}=\frac{r^2-ad}{r(a+d)}. \end{align*}$ Thus by the lemma,
$e=\frac{r^2-ad}{a+d}=\frac{\frac{abc+abd+acd+bcd}{a+b+c+d}-ad}{a+d}=\frac{abc+bcd-a^2d-ad^2}{(a+b+c+d)(a+d)}=\frac{bc-ad}{a+b+c+d},$ as desired.

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