MP = NQ wanted, incircles related

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30653 posts

#1 h Sep 22, 2020, 11:29 PM • 11 Y

Y by buratinogigle, nguyendangkhoa17112003, Nathanisme, megarnie, ImSh95, MightyDog, PHSH, deplasmanyollari, Rounak_iitr, Mathandski, NicoN9

Let $ABC$ be an acute-angled triangle and let $D, E$ , and $F$ be the feet of altitudes from $A, B$ , and $C$ to sides $BC, CA$ , and $AB$ , respectively. Denote by $\omega_B$ and $\omega_C$ the incircles of triangles $BDF$ and $CDE$ , and let these circles be tangent to segments $DF$ and $DE$ at $M$ and $N$ , respectively. Let line $MN$ meet circles $\omega_B$ and $\omega_C$ again at $P \ne M$ and $Q \ne N$ , respectively. Prove that $MP = NQ$ .

(Vietnam)

This post has been edited 1 time. Last edited by parmenides51, Sep 22, 2020, 11:30 PM

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pad

1671 posts

pad

#2 h Sep 22, 2020, 11:30 PM • 3 Y

Y by Nathanisme, ImSh95, NicoN9

Diagram

Let $I_1,I_2$ denote the centers of $\omega_b,\omega_c$ respectively, and let $N',M'$ denote the feet from $I_1,I_2$ to $BC$ respectively.

Lemma: $\angle NM'Q = \angle MND$ .
Proof: $\angle NM'Q = \tfrac12 \widehat{QN} = \angle QNE = \angle MND$ . $\blacksquare$

Note that $\angle NQM'=\tfrac12 \widehat{M'N} = \angle DNM'=90-A/2$ . Now, by LoS on $\triangle NQM'$ , $\begin{align*} &\qquad \frac{NQ}{\sin \angle NM'Q} = \frac{M'N}{\sin \angle NQM'} \implies \frac{NQ}{\sin \angle MND} = \frac{M'N}{\sin (90-A/2)}\\ &\implies NQ = \frac{M'N\sin \angle MND}{\sin (90-A/2)}=\frac{M'N}{ND} \cdot \frac{ND\sin \angle MND}{\sin (90-A/2)}. \end{align*}$ Similarly, $MQ = \frac{MN'}{MD} \cdot \frac{MD\sin \angle NMD}{\sin(90-A/2)}.$ But $M'N/ND=MN'/MD$ since $\triangle DMN'\sim \triangle DNM'$ are similar iscoceles triangles, and $ND\sin \angle MND=MD\sin \angle NMD$ by LoS on $\triangle MND$ . Therefore, $MP=NQ$ .

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v_Enhance

6882 posts

v_Enhance

#3 h Sep 22, 2020, 11:32 PM • 21 Y

Y by Mathematicsislovely, Aryan-23, Nathanisme, srijonrick, Math_olympics, v4913, HamstPan38825, Boris2005, cttg8217, samrocksnature, ImSh95, akasht, a_n, Rounak_iitr, shafikbara48593762, Stuffybear, acuri, Mathandski, NicoN9, Captainscrubz, s27_SaparbekovUmar

Let $k$ be the ratio of the similarity between $\triangle DBF \sim \triangle DEC$ . Let $r_b$ and $r_c$ denote the inradii of these two triangles. Note that $\frac{MP}{NQ} = \frac{2r_b \sin \angle DMP}{2r_c \sin \angle DNQ} = \frac{r_b}{r_c} \div \frac{\sin \angle DMN}{\sin \angle DNM} = \frac{r_b}{r_c} \div \frac{DN}{DM}.$ Obviously $\frac{r_b}{r_c} = k$ . But $DN$ and $DM$ correspond to the distance from the vertex $D$ to the incircle, so $\frac{DN}{DM} = k$ as well. Thus $\frac{MP}{NQ} = k \div k = 1$ and we're done.

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jj_ca888

2725 posts

jj_ca888

#4 h Sep 22, 2020, 11:32 PM • 3 Y

Y by buratinogigle, Nathanisme, ImSh95

This was also Canada IMO TST # 2

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VulcanForge

626 posts

VulcanForge

#5 h Sep 23, 2020, 12:26 AM • 2 Y

Y by Nathanisme, ImSh95

Let the radii of $\omega_B, \omega_C$ be $r_1, r_2$ respectively. Also let the acute angles made by $(\overline{MN}, \overline{DF}),(\overline{MN},\overline{DE})$ be $\alpha, \beta$ . Then it suffices to note by extended law of sines that $\frac{r_1}{r_2} = \frac{DM}{DN} = \frac{\sin(\beta)}{\sin(\alpha)}$

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MP8148

888 posts

MP8148

#6 h Sep 23, 2020, 2:52 AM • 2 Y

Y by Nathanisme, ImSh95

Let $F_M$ , $F_N$ be the projections of $M$ , $N$ on $\overline{BC}$ ; $I_B$ , $I_C$ be the centers of $\omega_B$ , $\omega_C$ ; and suppose $\omega_B$ , $\omega_C$ are tangent to $\overline{BC}$ at $X$ , $Y$ . Note that $\angle CDE = \angle BDF = \angle A$ .

The desired statement is equivalent to proving $\text{Pow}(M,\omega_C) = \text{Pow}(N,\omega_B)$ . Denote $DM = DX = s_B$ and $DN = DY = s_C$ . We calculate $\begin{align*} \text{Pow}(N,\omega_B) &= NI_B^2 -I_BX^2 \\ &= (NF_N-I_BX)^2 + F_NX^2 - I_BX^2 \\ &= NF_N^2 - 2(NF_N)(I_BX) + F_NX^2 \\ &= NF_N^2 + (F_ND+DX)^2 - 2(DN \sin A)(DM \tan \tfrac{A}{2}) \\ &= (NF_N^2 + F_ND^2) + DX^2 + 2 (F_ND)(DX) - 2(DN \sin A)(DM \tan \tfrac{A}{2}) \\ &= DN^2 + DX^2 + 2(DN \cos A)(DX) - 2(DN \sin A)(DM \tan \tfrac{A}{2}) \\ &= s_C^2 + s_B^2 + 2s_Bs_C \cos A - 2s_Bs_C \sin A \tan \tfrac{A}{2},\end{align*}$ which is symmetric in $B$ and $C$ as desired.

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Aryan-23

558 posts

Aryan-23

#7 h Sep 23, 2020, 2:58 AM • 2 Y

Y by Nathanisme, ImSh95

Let the incircle of $\triangle BDF$ have radius $r_1$ and hit $BC$ at $D_1$ . Similarly assume that incircle of $\triangle DCE$ have radius $r_2$ and hit $BC$ at $D_2$ .

The key idea is to use the similarity $\triangle DBF \sim \triangle DEC$ . Note that this means : $\tfrac {DN}{DM}=\tfrac {r_1}{r_2}=\tfrac {\cos B}{\cos C}$ .

Let $\angle DMN=\angle FMP= \angle MD_1P=\theta$ and $\angle DNM=\angle ENQ= \angle ND_2Q=\varphi$

Since we have $MP=2r_1\cdot \sin \theta$ and $NQ=2r_2\cdot \sin \varphi$ . Hence we aim to prove: $\tfrac {\sin \theta}{\sin \varphi}=\tfrac{\cos C}{\cos B}$ .

Note that by law of sines in $\triangle DNM$ we have :

$\frac {\sin \theta}{\sin \varphi}=\frac {DN}{DM}=\frac{\cos C}{\cos B}$
This concludes the proof. $\blacksquare$

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lahmacun

259 posts

lahmacun

#8 h Sep 23, 2020, 9:26 AM • 3 Y

Y by Nathanisme, ImSh95, Mango247

Let $b,c$ be the inradii of $\triangle BDF$ and $\triangle CDE$ respectively. By chord length formula, it suffice to show that $\frac{b}{c}=\frac{\sin\angle QND}{\sin\angle PMD} \iff \frac{b}{c}=\frac{DM}{DN}$ which is true because $\triangle BDF\sim \triangle EDC$

This post has been edited 1 time. Last edited by lahmacun, Sep 23, 2020, 9:27 AM

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lilavati_2005

357 posts

lilavati_2005

#10 h Sep 23, 2020, 11:40 AM • 2 Y

Y by Nathanisme, ImSh95

$X = \omega_B \cap BC$
$Y =\omega_C \cap BC$

Note that :
$\angle FMP = \angle NMD = \angle MCP = \alpha$
$\angle ENQ = \angle MND = \angle NYQ = \beta$

By Law of Sines in $\triangle MPX, \triangle NQY$ and $\triangle DMN$ :
$\frac{MP}{\sin \alpha} = \frac{MX}{\cos A/2}$
$\frac{NQ}{\sin \beta} = \frac{NY}{\cos A/2}$
$\frac{DM}{DN} = \frac{\sin \alpha}{\sin \beta}$

$\begin{align*} &\frac{MP}{NQ}\\ &=\frac{MX}{NY}\cdot\frac{\sin \alpha}{\sin \beta}\\ &=\frac{MX}{NY}\cdot\frac{DN}{DM}\\ &=\frac{MX}{NY}\cdot\frac{DY}{DX}\\ &=\frac{\sin A}{\sin A/2}\cdot\frac{\sin A/2}{\sin A}\\ &=\boxed{1} \end{align*}$

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franchester

1487 posts

franchester

#11 h Sep 24, 2020, 4:32 AM • 5 Y

Y by Nathanisme, ImSh95, Mango247, Mango247, Mango247

was not expecting this to die so easily to trig lol (note to self: try dumb things first spoiler)

Let $\omega_B$ and $\omega_C$ intersect $BC$ at $X,Y$ respectively. Furthermore, define $r_b$ and $r_c$ as the radii of $\omega_B$ and $\omega_C$ , respectively. We want to show $2r_b\sin\angle PXM=2r_c\sin\angle QYN$ From basic angle chasing, we have that $2r_b\sin\angle PXM=2r_b\sin\angle PMF=2r_b\sin\angle DMN$ and similarly we get $2r_c\sin\angle QYN=2r_c\sin\angle DNM$ . Rearranging and applying Law of Sines, it suffices to show that $\frac{r_b}{r_c}=\frac{DM}{DN} \iff \frac{r_b}{DM}=\frac{r_c}{DN}$ However, this is immediate by the fact that $\triangle BFD\sim \triangle BCA\sim \triangle ECD$ , so we are done.

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thepsserby

18 posts

thepsserby

#12 h Sep 24, 2020, 10:04 PM • 3 Y

Y by Nathanisme, ImSh95, IAmTheHazard

Let $I_B$ and $I_C$ be the centres of $\omega_B$ and $\omega_C$ , and let $R$ and $S$ be their respective tangency points with $BC$ . Let $X$ be the midpoint of $MN$ . It suffices to show that $X$ lies on the radical axis of the two circles. Clearly the midpoint $Y$ of $RS$ lies on the radical axis, so it remains to prove that $XY\perp I_BI_C$ .

Set $D=(0,0)$ in the Cartesian plane with $BC$ being the $x$ -axis. Suppose $R=(-r,0)$ , $S=(s,0)$ and $\angle{RDI_B}=\angle{I_BDM}=\angle{NDI_C}=\angle{I_CDS}=\alpha$ . Then, it is easy to compute the following:
$I_B=(-r,r\tan{\alpha}) ,\hspace{5pt}I_C=(s,s\tan{\alpha}) ,\hspace{5pt}M=(-r\cos{2\alpha},r\sin{2\alpha}) ,\hspace{5pt}N=(s\cos{2\alpha},s\sin{2\alpha}) ,\hspace{5pt}X=\left(\frac 12(s-r)\cos{2\alpha},\frac 12(s+r)\sin{2\alpha}\right) ,\hspace{5pt}Y=\left(\frac 12 (s-r),0\right)$ Now we can check the product of the gradients of $XY$ and $I_BI_C$ :
$\frac{\frac 12 (s+r)\sin{2\alpha}}{\frac 12(s-r)(\cos{2\alpha}-1)}\times \frac{(s-r)\tan{\alpha}}{s+r}=\frac{\sin{2\alpha}\tan{\alpha}}{\cos{2\alpha}-1}=\frac{2\sin^2{\alpha}}{-2\sin^2{\alpha}}=-1$ Hence, the lines are perpendicular, as required.

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pi_quadrat_sechstel

610 posts

pi_quadrat_sechstel

#13 h Sep 26, 2020, 7:31 AM • 2 Y

Y by Nathanisme, ImSh95

Let $I_B$ and $I_C$ be the midpoints of $\omega_B$ and $\omega_C$ . Let $\omega_B$ and $\omega_C$ be tangent to $CB$ at $R$ and $S$ . Let $X$ and $Y$ be the midpoints of $RS$ and $NM$ .

Then $X$ lies on the power axis of $\omega_B$ and $\omega_C$ . There is a spiral similarity between $FB$ and $CE$ with center $D$ .

Use complex coordinates with $d=0$ and $i_B=1$ . We have $m=\overline{r}, n=ri_C, s=\overline{r}i_C, x=\frac{r+\overline{r}i_C}{2}$ and $y=\frac{\overline{r}+ri_C}{2}$ .

$\frac{y-x}{i_B-i_C}=\frac{r+\overline{r}i_C-\overline{r}-ri_C}{2(1-i_C)}=\frac{(r-\overline{r})(1-i_C)}{2(1-i_C)}=\frac{r-\overline{r}}{2}=i\Im(r)$
We get $XY\perp I_BI_C$ . $Y$ lies on the power axis of $\omega_B$ and $\omega_C$ .We get:

$\begin{align*} |YN|\cdot|YQ|=|YM|\cdot|YP|\\ |YN|\cdot(|YN|+|NQ|)=|YM|\cdot(|YM|+|MP|)\\ |YN|+|NQ|=|YM|+|MP|\\ |NQ|=|MP| \end{align*}$

This post has been edited 1 time. Last edited by pi_quadrat_sechstel, Sep 26, 2020, 7:34 AM
Reason: mistake in line 3

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mela_20-15

125 posts

mela_20-15

#14 h Sep 29, 2020, 2:55 PM • 6 Y

Y by Nathanisme, ImSh95, plagueis, Mango247, Mango247, Mango247

Let $BC$ be tangent at $w_B,w_C$ at points $P_B,P_C$ and the second common external tangent at $Q_B,Q_C$ .
Our first claim is that $Q_BQ_C$ is parallel to $EF$ .
Let $I_B,I_C$ be the incenters , the triangle $I_BDI_C$ is similar to $BDE$ because $\angle I_BDI_C=\angle BDE=180-\angle A$
and from similarity $\frac{DI_B}{DI_C}=\frac{BD}{DC}$ . We eventually we get that $\angle DI_BI_C=90-\angle C$ and $DI_CI_B=90-\angle B$ and that the angle between $BC,I_BI_C$ is $(\angle B -\angle C)/2$ and that of $BC,Q_BQ_C$ is $\angle B-\angle C$ and that $EF,Q_BQ_C$ are parallel.
Then $P_C,M,Q_B$ and $P_B,N,Q_C$ are collinear :
Again we ge that $MDP_C$ is similar to $BDE$ from where we get that $\angle DMP_C=90-\angle C$ and from $EF//Q_BQ_C$ we have that arc
$MQ_B = 90-\angle C$ thus $P_CM$ passes through $Q_B$ .
Let $P_CM$ meet $w_C$ at $R_C$ and $P_BN$ , $w_B$ at $R_B$ .
We have that $P_CR_C=MQ_B$ because $P_CM\cdot PQ_B=P_CP_B^2=Q_BQ_C^2=Q_BR_C\cdot Q_BP_C$ This shows that the radical axis of $w_B,w_C$ passes through the midpoints of $Q_BP_C$ and $Q_CP_B$ which in turn shows that it passes through the midpoint of $MN$ which is what we were looking for.

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amuthup

779 posts

amuthup

#15 h Oct 2, 2020, 2:13 AM • 5 Y

Y by Aryan-23, buratinogigle, albgeo, Nathanisme, ImSh95

Long synthetic solution.

Redefine $P,Q$ as the intersection points of $\omega_B,\omega_C$ with $\overline{BC}.$ Let $T=\overline{MQ}\cap\overline{NQ},$ and let $W,X,Y,Z$ denote the midpoints of $\overline{PM},\overline{MN},\overline{NQ},\overline{QP}$ respectively. Additionally, let $I_B$ and $I_C$ denote the incenters of $\omega_B$ and $\omega_C$ respectively.

$\textbf{Claim: }$ $PN=QM$

$\emph{Proof: }$ Note that there is spiral similarity centered at $D$ sending $\overline{BF}$ to $\overline{EC},$ so $\triangle DPN\sim\triangle DMQ.$ Therefore, since $DP=DM,$ we have $PN=QM,$ as desired. $\blacksquare$

$\textbf{Claim: }$ $\overline{DT}\perp\overline{I_{B}I_{C}}$

$\emph{Proof: }$ Since $D$ is the Miquel point of quadrilateral $PNMQ,$ the quadrilaterals $DTMP$ and $DTNQ$ are cyclic. But since $\angle DMI_{B}=\angle DPI_{B}=\angle DNI_{C}=\angle DQI_{C}=90^\circ,$ the quadrilaterals $DMI_{B}P$ and $DNI_{C}Q$ are cyclic. Hence, $\angle DTI_{B}=\angle DTI_{C}=90^\circ,$ implying the desired conclusion. $\blacksquare$

$\textbf{Claim: }$ $\overline{TD}$ bisects $\angle PTQ$

$\emph{Proof: }$ Since quadrilateral $DTI_{B}P$ is cyclic, $\angle DTP=\angle DI_{B}P=90^\circ-\frac{1}{2}\angle FDB.$ Similarly, we can show that $\angle DTQ=90^\circ-\angle EDC.$ But it is well-known that $\angle FDB=\angle EDC,$ so we are done. $\blacksquare$

$\textbf{Claim: }$ $\overline{XZ}\perp\overline{I_{B}I_{C}}$

$\emph{Proof: }$ Since $PN=QM,$ the Varignon parallelogram of quadrilateral $PMNQ$ is a rhombus, so $\overline{XZ}\perp\overline{WY}.$ Therefore, since $\overline{DT}\perp\overline{I_{B}I_{C}},$ it suffices to show that $\overline{WY}\parallel\overline{DT}.$

Let $G$ be the intersection point of the perpendicular bisectors of $\overline{PM}$ and $\overline{QN}.$ Since $GP=GM,GN=GQ,PN=MQ,$ we have $\triangle GPN\cong\triangle GMQ,$ so $G$ is the Miquel point of quadrilateral $PNMQ.$ Moreover, $G$ is equidistant from $\overline{PN}$ and $\overline{QM},$ so $G$ lies on $\overline{TD}.$

Note that by properties of spiral similarity, $\angle(\overline{WY},\overline{MQ})=\angle WGM.$ Thus, it suffices to show that $\angle GTQ+\angle WGM=90^\circ.$

Since $\angle GWM=90^\circ,$ we have $\angle WGM=90^\circ-\angle WMG=90^\circ - \angle PMG=90^\circ-\angle PTG=90^\circ-\angle GTQ,$ as desired. $\blacksquare$

Now remark that $Z$ lies on the radical axis of $\omega_B,\omega_C,$ so $\overline{XZ}$ is the radical axis of $\omega_B,\omega_C.$ This implies that $X$ has equal power with respect to both circles, which is exactly what we wanted to prove.

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r_ef

11 posts

r_ef

#16 h Oct 7, 2020, 6:46 AM • 4 Y

Y by ImSh95, Mango247, Mango247, Mango247

Let $I_1,I_2$ denote the centers of $\omega_b,\omega_c$ respectively, and let $T,K$ denote the feet from $I_1,I_2$ to $BC$ respectively.
I will prove $\text{Pow}(M,\omega_C) = \text{Pow}(N,\omega_B)$ . Its easy to see $NT=MK$ and $\angle I_1 TM =\angle ACB , \angle I_2 KM =\angle ABC$ , $\frac{I_1 T}{I_2 K}=\frac{\cos\angle ABC}{\cos\angle ACB}$
We know $NI_1 ^2 - I_1 T^2 =NT^2 - 2NT.I_1 T.\cos\angle I_1 TN = MK^2 - 2MK.I_2 K.\cos\angle I_2 KM = MI_2 ^2 - I_2 K^2$
We are done

This post has been edited 1 time. Last edited by r_ef, Oct 7, 2020, 6:48 AM
Reason: Edit

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john0512

4191 posts

john0512

#56 h Jan 18, 2024, 2:23 AM • 1 Y

Y by sophie.germain

WLOG $AC>AB$ (this doesn't really matter but it makes the explanation more smooth). The idea here is that $\omega_c$ is "larger" than $\omega_b$ , but the arc $NQ$ has a smaller measure than arc $MP$ to compensate. Thus we can essentially split the problem into two parts.

First, we find the ratio between the radiuses of $\omega_b$ and $\omega_c$ . Then, we find the ratio of the sines of half the arc measures, and show that these multiply to 1. This would solve the problem.

\begin{claim}
We have $\frac{R_{\omega_c}}{R_{\omega_b}}=\frac{\cos\gamma}{\cos\beta}.$ \end{claim}

Note that $ABDE$ is cyclic, so there is a homothety at $C$ followed by a reflection across the angle bisector that sends $\triangle ACB$ to $\triangle DCE$ . Since $\frac{CD}{AC}=\cos\gamma,$ this homothety has ratio $\cos\gamma$ , so $R_{\omega_c}=r\cos\gamma$ . Similarly, $R_{\omega_b}=r\cos\beta$ , which shows the claim.

Now, we find the "ratio of sines of half the arc measures" we were looking for. Since $NE$ is tangent to $\omega_c$ , this ratio is equal to $\frac{\sin\angle ENQ}{\sin\angle FMP}=\frac{\sin\angle DNM}{\sin\angle DMN}=\frac{DM}{DN}.$ However, $DM=\frac{DB+DF-FB}{2}=\frac{c\cos\beta+b\cos\beta-a\cos\beta}{2}=\cos\beta\cdot \frac{b+c-a}{2}.$ Similarly, $DN=\cos\gamma\cdot \frac{b+c-a}{2}.$ Thus, $\frac{DM}{DN}=\frac{\cos\beta}{\cos\gamma},$ which is the reciprocal of the ratio of radii, so we are done.

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ihatemath123

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ihatemath123

#57 h Feb 5, 2024, 1:21 AM

Y by

Unlike post #53 this is actually me now

Because $\angle FDB = \angle EDC$ , it follows that the ratio of the radius of $\omega_B$ to the radius of $\omega_C$ is $\frac{MD}{DN}$ . Now, letting $X$ and $Y$ be the tangency points of $\omega_B$ and $\omega_C$ with $\overline{BC}$ , it follows that
$\frac{MP}{NQ} = \frac{\text{radius} (\omega_B) \cdot \sin ( \angle MXP )}{\text{radius} (\omega_C) \cdot \sin ( \angle NYQ )} = \frac{MD}{DN} \cdot \frac{\sin ( \angle DMN )}{\sin (\angle DNM)} = 1.$
remark

This post has been edited 3 times. Last edited by ihatemath123, Feb 7, 2024, 6:07 AM

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Shreyasharma

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Shreyasharma

#58 h Feb 7, 2024, 4:31 AM

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draw((-0.4720251699614325,-1.593345575866262)--(-1.156553700140246,-1.3716452021444483), linewidth(0.7) + blue); draw((-0.4720251699614325,-1.593345575866262)--(-0.4176549586609028,-0.8758678797249145), linewidth(0.7) + blue); draw((-0.4720251699614325,-1.593345575866262)--(-0.4779470667117702,-2.3128560312983137), linewidth(0.7) + blue); draw((-2.0886459647031916,-1.8347001573988362)--(-2.0924720355047097,-2.2995677599551367), linewidth(0.7) + blue); draw((-2.3829807767200273,-2.194538378213741)--(-1.6440820352406826,-1.6987610557942068), linewidth(0.7) + ccqqqq); draw((-0.4176549586609028,-0.8758678797249145)--(-1.156553700140246,-1.3716452021444483), linewidth(0.7) + ccqqqq); draw((-1.9741056144780595,-0.6194793960715617)--(-0.7926360753077117,-0.2480025458385933), linewidth(0.7) + ccqqqq); /* dots and labels */ dot((-1.436498872725946,0.40140832631925505),dotstyle); label("$A$", (-1.4931395418506677,0.48825897414250724), NE * labelscalefactor); dot((-2.8555447851977687,-2.2932873258036057),dotstyle); label("$B$", (-3.1504944029878994,-2.407846216918973), NE * labelscalefactor); dot((1.2688308139529472,-2.327232803973365),dotstyle); label("$C$", (1.2975399794759466,-2.4388537671444706), NE * labelscalefactor); dot((-1.4471850279001193,-0.8969595278228518),linewidth(4pt) + dotstyle); label("$H$", (-1.7413561523719848,-0.7706475650126973), NE * labelscalefactor); dot((-1.4587720556310186,-2.3047833976476415),linewidth(4pt) + dotstyle); label("$D$", (-1.4807365217558826,-2.6450552771895704), NE * labelscalefactor); dot((-0.7926360753077117,-0.2480025458385933),linewidth(4pt) + dotstyle); label("$E$", (-0.7675628663057479,-0.20010864086354052), NE * labelscalefactor); dot((-1.9741056144780595,-0.6194793960715617),linewidth(4pt) + dotstyle); label("$F$", (-2.1194920566373074,-0.5721992435695124), NE * labelscalefactor); dot((-2.0886459647031916,-1.8347001573988362),linewidth(4pt) + dotstyle); label("$X$", (-2.1070890365425226,-1.750486152138423), NE * labelscalefactor); dot((-0.4720251699614325,-1.593345575866262),linewidth(4pt) + dotstyle); label("$Y$", (-0.4264798136991617,-1.7532770918678257), NE * labelscalefactor); dot((-1.6440820352406826,-1.6987610557942068),linewidth(4pt) + dotstyle); label("$M$", (-1.7357742729406946,-1.4954484010109347), NE * labelscalefactor); dot((-1.156553700140246,-1.3716452021444483),linewidth(4pt) + dotstyle); label("$N$", (-1.1334519591019039,-1.6458363206501384), NE * labelscalefactor); dot((-2.3829807767200273,-2.194538378213741),linewidth(4pt) + dotstyle); label("$P$", (-2.528791719765211,-2.1466069557463855), NE * labelscalefactor); dot((-0.4176549586609028,-0.8758678797249145),linewidth(4pt) + dotstyle); label("$Q$", (-0.682293404125694,-0.8416551152381949), NE * labelscalefactor); dot((-0.4779470667117702,-2.3128560312983137),linewidth(4pt) + dotstyle); label("$T_C$", (-0.5070994443152639,-2.663659807324869), NE * labelscalefactor); dot((-2.0924720355047097,-2.2995677599551367),linewidth(4pt) + dotstyle); label("$T_B$", (-2.1194920566373074,-2.657458297279769), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let the center of $\omega_B$ be $X$ and the center of $\omega_C$ be $Y$ . Also let $T_B$ and $T_C$ be the tangency points of $\omega_B$ and $\omega_C$ to $\overline{BC}$ . Note that we have,
$\begin{align*} \frac{MP}{NQ} &= \frac{2r_1\cos(\angle XMP)}{2r_2\cos(\angle YNQ)}\\ &= \frac{r_1\cos(90 - \angle PMD)}{r_2 \cos(90 - \angle QNE)}\\ &= \frac{r_1 \sin(\angle PMD)}{r_2 \sin(\angle QNE)}\\ &= \frac{r_1 \sin(180 - \angle NMD)}{r_2 \sin(180 - \angle MND)}\\ &= \frac{r_1 \sin(\angle NMD)}{r_2 \sin (\angle MND)}\\ &= \frac{r_1 \cdot ND}{r_2 \cdot MD}\\ &= \frac{r_1 \cdot r_2 \cdot \tan(\angle NYD)}{r_1 \cdot r_2 \cdot \tan(\angle MXD)}\\ &= \frac{\tan(\angle DYT_C)}{\tan(\angle DXT_B)}\\ &= \frac{\tan(\angle XDT_B)}{\tan(\angle YDT_C)}\\ &= \frac{\tan\left(\frac{1}{2} \angle FDB \right)}{\tan\left( \frac{1}{2} \angle CDE \right)}\\ \end{align*}$ However note that,
$\begin{align*} \angle FDB &= 180 - \angle FDC\\ &= \angle A\\ &= 180 - \angle EDB\\ &= \angle EDC \end{align*}$ and hence the ratio is simply $1$ , as desired. $\square$

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shendrew7

806 posts

shendrew7

#59 h Mar 17, 2024, 3:40 PM

Y by

Let $r_b$ , $r_c$ be the radii of $\omega_b$ , $\omega_c$ . Thus we have
$\frac{MP}{NQ} = \frac{r_b \cdot \sin (180-\angle DMN)}{r_c \cdot \sin (180-\angle DNM)} = \frac{r_1/r_2}{DM/DN} = 1$
from Law of Sines and the similarity $\triangle DFB \sim \triangle DCE$ . $\blacksquare$

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trk08

614 posts

trk08

#60 h Mar 22, 2024, 8:49 PM

Y by

Let $r_{B,C}$ and $O_{B,C}$ be the radii and center, respectively, of $\omega_{B,C}$ , respectively. Note that, by LoS:
$MP=2r_B\cos{\angle PMO_B}=2r_B\sin{\angle NMD}$ $NQ=2r_C\cos{\angle QNO_C}=2r_C\sin{\angle MND}$
Note that:
$r_B=MD\tan{\angle BDF/2}$ $r_C=DN\tan{\angle CDE/2}.$ By angle chasing on cyclic quads $BDHF$ , $CDHE$ ( $H$ is orthocenter):
$\angle BDF=\angle BHF=\angle CHE=\angle CDE.$ Therefore, using LoS:
$r_B/r_C=MD/DN=\sin{\angle MND}/\sin{\angle NMD}.$
Therefore:
$MP/NQ=1$ $\square$

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fearsum_fyz

56 posts

fearsum_fyz

#62 h Jun 23, 2024, 8:03 AM • 1 Y

Y by GeoKing

Let the incenters of $\Delta{BFD}$ and $\Delta{CED}$ be $I_1$ and $I_2$ and their inradii be $r_1$ and $r_2$ respectively. Then
$\frac{MP}{NQ} = \frac{2 r_1 \sin{\widehat{MP}}}{2 r_2 \sin{\widehat{NQ}}} \overset{\text{(Alt Segt Thm)}}{=} \frac{r_1 \sin{\angle{DMN}}}{r_2 \sin{\angle{DNM}}} \overset{\text{(LOS)}}{=} \frac{r_1 DN}{r_2 DM} \overset{\Delta{DI_1M} \overset{\text{(AA)}}{\sim} \Delta{DI_2N} }{=} 1$ as desired.

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Reason: then

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Ritwin

158 posts

Ritwin

#63 h Sep 9, 2024, 1:40 AM

Y by

trig? more like complex numbers.

Solution

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Markas

150 posts

Markas

#64 h Sep 22, 2024, 10:01 PM

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Let $I_1$ be the incenter in $\triangle BFD$ and $I_2$ be the incenter in $\triangle DEC$ . Let $I_1M = I_1P = r_b$ and $I_2N = I_2Q = r_c$ . Also let $\angle PI_1M = 2x$ and $\angle NI_2Q = 2y$ , where obviously $\angle I_1MP = \angle I_1PM = 90 - x$ and $\angle I_2NQ = \angle I_2QN = 90 - y$ . Also by some angle chase we get that $\angle PMD = x$ , $\angle DMN = 180 - x$ , $\angle DNM = y$ . Now by law of sines on $\triangle MPI_1$ we get that $MP = \frac{\sin \angle PI_1M \cdot I_1M}{\sin \angle I_1MP} = \frac{\sin 2x \cdot r_b}{\sin (90 - x)} = \frac{2\sin x \cdot \cos x \cdot r_b}{\cos x} = 2\sin x \cdot r_b = 2\sin (180 - x) \cdot r_b = 2\sin \angle DMN \cdot r_b$ . Now by law of sines on $\triangle NQI_2$ we get that $NQ = \frac{\sin \angle NI_2Q \cdot NI_2}{\sin \angle NQI_2} = \frac{\sin 2y \cdot r_c}{\sin (90 - y)} = \frac{2\sin y \cdot cos y \cdot r_c}{\cos y} = 2\sin y \cdot r_c = 2\sin \angle DNM \cdot r_c$ $\Rightarrow$ $\frac{MP}{NQ} = \frac{2\sin \angle DMN \cdot r_b}{2\sin \angle DNM \cdot r_c} = \frac{\sin \angle DMN \cdot r_b}{\sin \angle DNM \cdot r_c}$ . Now by law of sines on $\triangle DNM$ we get that $\frac{\sin \angle DMN}{\sin \angle DNM} = \frac{DN}{DM}$ $\Rightarrow$ $\frac{MP}{NQ} = \frac{DN}{DM} \cdot \frac{r_b}{r_c}$ . Also $\triangle DBE \sim \triangle DEC$ $\Rightarrow$ $\triangle DI_1M \sim \triangle DI_2N$ $\Rightarrow$ $\frac{DN}{DM} = \frac{r_c}{r_b}$ $\Rightarrow$ $\frac{MP}{NQ} = \frac{DN}{DM} \cdot \frac{r_b}{r_c} = \frac{r_c}{r_b} \cdot \frac{r_b}{r_c} = 1$ $\Rightarrow$ MP = NQ and we are ready.

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Mathandski

774 posts

Mathandski

#65 h Oct 16, 2024, 7:22 PM

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$\text{[math]}$

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SimplisticFormulas

131 posts

SimplisticFormulas

#66 h Dec 9, 2024, 4:03 PM

Y by

Okayy let’s bashh!
We shall show that $Pow_{\omega_{C}}(M)=Pow_{\omega_{B}}(N)$ (which is equivalent to proving the problem statement).
Let $r$ be the inradius of $\triangle ABC$ . Let $O_{1}$ be the incentre of $\triangle CED$ . Let the foot of perpendicular from $O_{1},M$ on $BC$ be $X,Y$ respectively.
Observe that $\triangle DBF \sim \triangle ABC \sim \triangle DEC$ .
Therefore, (skipping a few details) we get
$DM=(s-a)cosC=DX, DN=(s-a)cosB,DY=DMcosA=cosAcosB(s-a)$ , $MY=MDsinA=sinAcosB(s-a),O_{1}X=rcosC$
Hence, $Pow_{\omega_{C}}(M)=MO_{1}^2-(rcosC)^2$ $=(s-a)^2(cosAcosB+cosC)^2 + (rcosC-sinAcosB(s-a))^2-r^2cos^2C$ $=(s-a)^2cos^2Acos^2B+(s-a)^2cos^2C+2cosAcosBcosC(s-a)+(s-a)^2sin^2Acos^2B-2r(s-a)sinAcosBcosC$ $=(s-a)^2(cos^2Acos^2B+cos^2C+sin^2Acos^2B)+2(s-a)cosBcosC(cosA-rsinA)$ $=(s-a)^2(cos^2B+cos^2C)+ 2(s-a)cosBcosC(cosA-rsinA)$ which is symmetric in $B$ and $C$ , so we are done. $\blacksquare$

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maths_enthusiast_0001

133 posts

maths_enthusiast_0001

#67 h Jan 17, 2025, 9:04 PM • 1 Y

Y by L13832

Very nice and simple :-D

Claim: $\color{blue}{MP=NQ}$
Proof: Note that we have, $BF=a \cos B,FD=b \cos B,DB=c \cos B$ and $\angle{FDB}=\angle {A}$ . Also, $EC=a \cos C,CD=b \cos C, DE=c \cos C$ and $\angle{EDC}=\angle {A}$ . Clearly, $\Delta BFD \sim \Delta ECD$ . Denote the inradii of $\Delta BFD$ and $\Delta ECD$ by $r_{1}$ and $r_{2}$ respectively. Thus, $\boxed{\frac{r_1}{r_2}=\frac{\cos B}{\cos C}}$ . Let the incircles of triangles $BDF$ and $CDE$ be tangent to $BC$ at $X$ and $Y$ respectively. By alternate segment theorem, $\angle PXM=\alpha=\angle{PMF}=\angle{DMN}$ and $\angle{QYN}=\beta=\angle{QNE}=\angle{DNM}$ . Denote the semi-perimeter of $\Delta ABC$ by $s$ . Now note that, $DM=(s-a)\cos B$ and $DN=(s-a)\cos C$ implying, $\boxed{\frac{DM}{DN}=\frac{\cos B}{\cos C}}$ . Thus by sine rule in triangle $DMN$ we have,
$\frac{DM}{DN}=\frac{\sin \angle DNM}{\sin \angle DMN}=\frac{\sin \beta}{\sin \alpha} \implies \boxed{\frac{\sin \beta}{\sin \alpha}=\frac{\cos B}{\cos C}}$ Now note that $MP=2r_{1}\sin \alpha$ and $NQ=2r_{2}\sin \beta$ implying, $\frac{NQ}{MP}=\left(\frac{r_{2}}{r_{1}}\right)\left(\frac{\sin \beta}{\sin \alpha}\right)=\frac{\cos C.\cos B}{\cos B.\cos C}=1 \implies \boxed{MP=NQ}$ as desired. $\blacksquare$ ( $\mathcal{QED}$ )

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AngeloChu

471 posts

AngeloChu

#68 h Jan 22, 2025, 5:09 PM

Y by

let the points of tangency of $\omega_B$ and $\omega_C$ to $BC$ be $R$ and $S$ respectively
a simple angle chase yields that $MPR=NQS=MDN$ and $PMR=QSN$ , and $FMP=MND$
thus, $MPR$ is similar to $NMD$ and $NSQ$
then more angle chase gives that $DMR$ is similar to $DSN$
then, our first similarity condition yields $MP*DN/DM*NQ/QS=NQ$ , but our second condition yields $DN/DM=QS/NQ$ so $MP=MP*DN/DM*DM/DN=NQ$

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mananaban

36 posts

mananaban

#70 h Apr 17, 2025, 1:28 AM • 1 Y

Y by buratinogigle

My first instinct was trying to prove $PM \cdot MQ = QN \cdot NP$ or $PM \cdot MN = QN \cdot NM$ with some synthetic power stuff, but alas. Kudos to #20 for the power solution!

Let $r(\omega)$ denote the radius of circle $\omega$ .
First observe the easy similarity chain of $\triangle BDF \sim \triangle BAC \sim \triangle EDC$ . Note how $r(\omega_B) / r(\omega_C)$ is the scale factor between $\triangle BDF \sim \triangle EDC$ , which can be expressed as $DM/DN$ .

Now the proof is an easy trigbash/lengthchase. Note that the first step below is due to the angle intercepted by $MP,NQ$ on $\omega_B,\omega_C$ being equal to $\angle PMF,\angle ENQ$ respectively.
$\begin{align*} \frac{MP}{NQ} &= \frac{r(\omega_B) \cdot \sin(\angle PMF)}{r(\omega_C) \cdot \sin(\angle ENQ)} \\ &= \frac{r(\omega_B)}{r(\omega_C)} \cdot \frac{\sin(\angle DMN)}{\sin(\angle DNM)} \\ &= \frac{DM}{DN} \cdot \frac{DN}{DM}, \end{align*}$ which is $1$ . Thus, $MP=NQ$ , as desired. $\blacksquare$

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Giant_PT

54 posts

Giant_PT

#71 h Apr 17, 2025, 7:27 AM

Y by

Let $M'$ and $N'$ be the intersection between line $BC$ and circles $\omega_B$ and $\omega_C$ respectively. Also, let $r_B$ and $r_C$ be lengths of the radius of circles $\omega_B$ and $\omega_C$ respectively.

Since we have that $\triangle ABC \sim \triangle DBF \sim \triangle DEC$ , we can calculate that $\frac{r_B}{r_C}=\frac{cosB}{cosC}$ and also that $MD=cosB\frac{(b+c-a)}{2}$ and $ND=cosC\frac{(b+c-a)}{2}.$ Therefore, $\frac{MD}{ND}=\frac{cosB}{cosC}$ . Via angle chasing, we have that $\angle QN'N = \angle QNE = \angle MND$ and $\angle PM'M = \angle PMF = \angle DMN$ .

Now, calculating ratios using los a few times, we have,
$\frac{cosB}{cosC}=\frac{r_B}{r_C}=\frac{\frac{MP}{2sin\angle PM'M}}{\frac{NQ}{2sin \angle QN'N}}=\frac{MP}{NQ} \cdot \frac{sin\angle QN'N}{sin\angle PM'M}=\frac{MP}{NQ} \cdot \frac{sin\angle MND}{sin\angle DMN}=\frac{MP}{NQ} \cdot \frac{MD}{ND}=\frac{MP}{NQ} \cdot \frac{cosB}{cosC},$ which clearly implies $\frac{MP}{NQ}=1$ as desired.

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ezpotd

1324 posts

ezpotd

#72 h May 26, 2025, 6:28 PM

Y by

The measure of arc $MP$ is $\angle FMP = \angle NMD$ , so we have (using $BDE \sim BAC$ with a factor of $\cos B$ ) $MP= r\cos B \sin NMD$ , symmetrically $NQ = r \cos C \sin MND$ , so it remains to prove $\frac{\sin MND}{\sin NMD}= \frac{\cos B}{\cos C}$ , by LoS it is sufficient to prove $\frac{DM}{DN} = \frac{\cos B}{\cos C}$ , which follows directly from the previous similarity, $MD = (s - a) \cos B, ND = (s - a) \cos C$ .

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