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Slightly weird points which are not so weird

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Source: India TST 2023 Day 4 P1

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#1 h Jul 9, 2023, 2:27 PM • 1 Y

Y by GeoKing

Suppose an acute scalene triangle $ABC$ has incentre $I$ and incircle touching $BC$ at $D$ . Let $Z$ be the antipode of $A$ in the circumcircle of $ABC$ . Point $L$ is chosen on the internal angle bisector of $\angle BZC$ such that $AL = LI$ . Let $M$ be the midpoint of arc $BZC$ , and let $V$ be the midpoint of $ID$ . Prove that $\angle IML = \angle DVM$

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#2 h Jul 9, 2023, 6:02 PM • 1 Y

Y by GeoKing

Trig

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#3 h Jul 9, 2023, 6:52 PM • 2 Y

Y by GeoKing, Ywgh1

Solution

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#4 h Jul 10, 2023, 10:28 AM • 1 Y

Y by GeoKing

Let $N$ be the other midpoint of arc $BC$ . Let $S$ be the Sharky Devil point (i.e., $MD\cap ZI$ ), and construct the rectangle $ANPI$ . Then, notice that
$\left. \begin{array}{l} \measuredangle DIM = \measuredangle(\perp BC, AM) = \measuredangle MNZ \\[4pt] \measuredangle IMD = \measuredangle AMS = \measuredangle AZS = \measuredangle AZI = \measuredangle PMN \end{array} \right\} \implies \triangle IMD\sim\triangle NMP$ with corresponding midpoints $V$ and $L$ . This implies the problem.

This post has been edited 1 time. Last edited by MarkBcc168, Jul 10, 2023, 10:28 AM

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#5 h Jul 20, 2023, 3:32 AM

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Let the angle bisector of $\angle BZC$ meet $\odot ABC$ at $M'$ . Let $X$ denote the midpoint of $AI$ . Let $O$ denote the centre of $\odot ABC$ . Thus, $AI \perp XL$ , and $MOM'$ are collinear.

Claim: $AMZM'$ is a parallelogram
Proof: As $AO=OZ=OM=OM'$ , so the diagonals bisect each other, which implies the claim.

Claim: $AXLM'$ is a parallelogram
Proof: $\angle AM'Z=\angle XAM'=\angle AXL=90^{\circ}$ . So, $AM' \parallel XL$ . By the previous claim, $AX \parallel M'L$

So, we get that $M'L=AX=\frac{AI}{2}$

Claim: $\triangle IVM \sim \triangle M'LM$
Proof: $\angle MM'Z=\angle MAZ=\angle MAC-\angle OAC=\angle BAM-\angle BAF=\angle FAM=\angle DIM$ . Also, $\frac{IV}{M'L}=\frac{r}{AI}$ where $r$ denotes the inradius. Then again, by the well known identity $AI . IM=2Rr$ , we get, $\frac{r}{AI}=\frac{IM}{2R}=\frac{IM}{MM'}$ These two facts imply the claim.

Now, summing up, $\angle IML=\angle MLZ=\angle LM'M+\angle LMM'=\angle VIM+\angle VMI=\angle DVM$ $\blacksquare$

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#6 h Oct 9, 2023, 12:40 AM

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Let $N$ be the midpoint of arc $BAC$ (also the $M$ -antipode), so $\overline{ZN}$ is the internal bisector of $\angle BZC$ . Let $K$ be the reflection of $N$ over $L$ , which is also the foot of the perpendicular from $I$ to $\overline{ZN}$ . Since $\angle IML=\angle KLM$ , it clearly suffices to show $\triangle IDM \sim \triangle NKM$ . Let $H'$ be the second intersection of the $A$ -altitude with $(ABC)$ . Since $M$ is well-known to be the midpoint of arc $H'Z$ , it follows that $\measuredangle MNK=\measuredangle H'AM=\measuredangle DIM$ , so it suffices to show $\frac{ID}{IM}=\frac{NK}{NM}$ . Since $AIKN$ is a rectangle, this is equivalent to $AI\cdot IM=2Rr$ where $r,R$ are the inradius and circumradius of $\triangle ABC$ respectively. This is well-known to be true, since $AI\cdot IM=\mathrm{Pow}_{(ABC)}(I)=OI^2-R^2$ . $\blacksquare$

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#7 h Feb 26, 2025, 5:13 AM

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A bad solution to a nice problem. We denote by $U$ the midpoint of segment $AI$ , by $P$ the foot of the $A-$ altitude and by $N$ the midpoint of segment $AP$ . Let $I_a$ denote the $A-$ excenter.

Note that by the Midpoint Theorem it follows that $DI_a \parallel VM$ . Also, the Midpoint of the Altitude Lemma tells us that points $N$ , $D$ and $I_a$ are collinear. Let $M'$ denote the major $BC$ arc midpoint. We start off by noting that since $N'$ is the major $BC$ arc midpoint, $\overline{ZL}$ must pass through $M'$ . Thus, it suffices to show the following claim.

Claim : Triangles $\triangle NPD$ and $\triangle MLU$ are similar.

Proof : First observe that since $LA=LI$ point $L$ must lie on the perpendicular bisector of segment $AI$ . Thus, $AI \perp UL$ . Further, due to diameters $AZ$ and $MN$ it also follows that $AM \perp MZ$ and $MZ \perp M'L$ . Thus, quadrilateral $ULZM$ is a rectangle.

Let $X$ be the foot of the $A-$ angle bisector on side $BC$ . Now note that,
$\frac{MZ}{AM}=\frac{PX}{AP}$ and since we know that $AM \cdot AX = AB \cdot AC$ and $AP \parallel DI$ ,
$\frac{MZ\cdot AP}{AB \cdot AC}=\frac{PX}{AP}= \frac{PD}{AI}$ Also we know $AI_a\cdot AI = AB \cdot AC$ . Thus,
$\frac{UL}{UM} = \frac{MZ}{UM} = \frac{2MZ}{AI_a} = \frac{2 PD \cdot AB \cdot AC \cdot AI}{AI \cdot AP \cdot AB \cdot AC} = \frac{2PD}{AP}=\frac{PD}{NP}$ which in conjunction with $\angle NPD = 90^\circ = \angle LUM$ is sufficient to imply the claim.

Now this means,
$\measuredangle UML = \measuredangle DNP = \measuredangle I_aNP = \measuredangle I_aDI = \measuredangle MVI = \measuredangle MVD$ which is the desired result.

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#8 h Feb 26, 2025, 5:21 PM

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Denote $I_a, I_b, I_c$ the corresponding excenters, $K$ the touch point of $(I_a)$ and $BC$ , $T$ the midpoint of $BC$ , $J$ the midpoint of segment $AI$ and $N$ the midpoint of arc $BAC$ .

Notice that $\triangle I_aAN \sim \triangle I_aKT$ by similarity between $\triangle I_aBC$ and $\triangle I_aI_bI_c$ . We also know through a common result that $KT = DT$ , thus:
$\cfrac{I_aK}{TK} = \cfrac{I_aA}{AN} \implies \cfrac{I_aK}{DK} = \cfrac{MJ}{AN}$ Then, as $ANZM$ and $JLZM$ are rectangles, it follows that $\cfrac{MJ}{AN} = \cfrac{MJ}{LJ}$
Henceforth, $\triangle I_aKD \sim \triangle MJL$ , and we finish off the problem with an angle chase:
$\angle IML = \angle I_aML = \angle KI_aD = 180^{\circ} - \angle IDJ = \angle DVM$

This post has been edited 4 times. Last edited by Retemoeg, Mar 18, 2025, 4:58 PM

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#9 h Mar 18, 2025, 11:50 AM • 1 Y

Y by Retemoeg

Retemoeg wrote:

Denote $I_a, I_b, I_c$ the corresponding excenters, $K$ the touch point of $(I_a)$ and $BC$ , $T$ the midpoint of $BC$ , $J$ the midpoint of segment $AI$ and $N$ the midpoint of arc $BAC$ .

Notice that $\triangle I_aAN \sim \triangle I_aKT$ by similarity between $\triangle I_aBC$ and $\triangle I_aI_bI_c$ . We also know through a common result that $KT = DT$ , thus:
$\cfrac{I_aK}{TK} = \cfrac{I_aA}{AN} \implies \cfrac{I_aK}{DK} = \cfrac{MJ}{AN}$ Then, as $ANZM$ and $JLZM$ are rectangles, it follows that $\cfrac{MJ}{AN} = \cfrac{MJ}{LJ}$
Henceforth, $\triangle I_aKD \sim \triangle MJL$ , and we finish off the problem with an angle chase:
$\angle IML = \angle JML = \angle KJD = 180^{\circ} - \angle IDJ = \angle DVM$

Thanks for good solution based on deep facts about similarity involving excenters as well. Just a friendly comment that there is a small typo; in the final angle chase I think the second and third $J$ should change to $I_a$ is it right?

This post has been edited 1 time. Last edited by kes0716, Mar 18, 2025, 11:52 AM

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#10 h Mar 18, 2025, 4:58 PM

Y by

kes0716 wrote:

Retemoeg wrote:

Denote $I_a, I_b, I_c$ the corresponding excenters, $K$ the touch point of $(I_a)$ and $BC$ , $T$ the midpoint of $BC$ , $J$ the midpoint of segment $AI$ and $N$ the midpoint of arc $BAC$ .

Notice that $\triangle I_aAN \sim \triangle I_aKT$ by similarity between $\triangle I_aBC$ and $\triangle I_aI_bI_c$ . We also know through a common result that $KT = DT$ , thus:
$\cfrac{I_aK}{TK} = \cfrac{I_aA}{AN} \implies \cfrac{I_aK}{DK} = \cfrac{MJ}{AN}$ Then, as $ANZM$ and $JLZM$ are rectangles, it follows that $\cfrac{MJ}{AN} = \cfrac{MJ}{LJ}$
Henceforth, $\triangle I_aKD \sim \triangle MJL$ , and we finish off the problem with an angle chase:
$\angle IML = \angle JML = \angle KJD = 180^{\circ} - \angle IDJ = \angle DVM$

Thanks for good solution based on deep facts about similarity involving excenters as well. Just a friendly comment that there is a small typo; in the final angle chase I think the second and third $J$ should change to $I_a$ is it right?

Yup, my bad I've changed it, thanks!

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#11 h Mar 19, 2025, 10:27 AM

Y by

Retemoeg wrote:

kes0716 wrote:

Retemoeg wrote:

Denote $I_a, I_b, I_c$ the corresponding excenters, $K$ the touch point of $(I_a)$ and $BC$ , $T$ the midpoint of $BC$ , $J$ the midpoint of segment $AI$ and $N$ the midpoint of arc $BAC$ .

Notice that $\triangle I_aAN \sim \triangle I_aKT$ by similarity between $\triangle I_aBC$ and $\triangle I_aI_bI_c$ . We also know through a common result that $KT = DT$ , thus:
$\cfrac{I_aK}{TK} = \cfrac{I_aA}{AN} \implies \cfrac{I_aK}{DK} = \cfrac{MJ}{AN}$ Then, as $ANZM$ and $JLZM$ are rectangles, it follows that $\cfrac{MJ}{AN} = \cfrac{MJ}{LJ}$
Henceforth, $\triangle I_aKD \sim \triangle MJL$ , and we finish off the problem with an angle chase:
$\angle IML = \angle JML = \angle KJD = 180^{\circ} - \angle IDJ = \angle DVM$

Thanks for good solution based on deep facts about similarity involving excenters as well. Just a friendly comment that there is a small typo; in the final angle chase I think the second and third $J$ should change to $I_a$ is it right?

Yup, my bad I've changed it, thanks!

I think it is
$\angle IML = \angle JML = \angle KI_aD = 180^{\circ} - \angle IDI_a = \angle DVM$

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