Hard to approach it !

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BogG

#1 h May 25, 2006, 9:45 PM • 15 Y

Y by malcolm, esi, Viswanath, steppewolf, pog, Ya_pank, HWenslawski, Adventure10, jhu08, megarnie, ImSh95, Mango247, Zhaom, Rounak_iitr, and 1 other user

Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$ . Let $H$ be the orthocenter of triangle $ABC$ , and let $M$ be the midpoint of the side $BC$ . Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$ , $H$ , $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$ .

This post has been edited 4 times. Last edited by BogG, May 26, 2006, 2:18 PM

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TomciO

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TomciO

#2 h May 25, 2006, 10:32 PM • 7 Y

Y by nguyendangkhoa17112003, pog, Adventure10, jhu08, megarnie, ImSh95, and 1 other user

"Common twist" = radical axis in this case I think.

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BogG

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BogG

#3 h May 25, 2006, 10:35 PM • 5 Y

Y by pog, Adventure10, jhu08, megarnie, ImSh95

I'm sorry I don't know the word. This is in this case for exemple the line of power of the two circles... Do you understand ?

"corde commune" in French...
"coarda comuna" in Rumania...

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Virgil Nicula

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Virgil Nicula

#4 h May 27, 2006, 7:44 PM • 6 Y

Y by pog, jhu08, ImSh95, Adventure10, Mango247, and 1 other user

Here is a syntetical proof of this nice and difficult problem. See PP12 from here

This post has been edited 9 times. Last edited by Virgil Nicula, Apr 3, 2016, 4:10 PM

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Megus

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Megus

#5 h May 28, 2006, 8:42 AM • 6 Y

Y by pog, Adventure10, TurtleKing123, jhu08, ImSh95, Mango247

Could you explain why $AA_3$ is an angle bisector of $OAH$ and why $A', M,H$ are collinear?

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Virgil Nicula

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Virgil Nicula

#6 h May 28, 2006, 6:07 PM • 6 Y

Y by pog, Adventure10, jhu08, megarnie, ImSh95, Mango247

Megus wrote:

Could you explain why $AA_3$ is an angle bisector of $OAH$ and why $A',M,H$ are collinear?

Lemma. Let $w=C(O)$ be a circumcircle of the acute triangle $ABC$ , where $AB<AC$ . Denote the point $A'\in w\cap (AO$ . Define the points $X\in AB$ , $Y\in AC$ so that $B\in (AX)\ ,\ AX=AC$ and $Y\in (AC)\ ,\ AY=AB$ . Then $m(\widehat {A'BY})=m(\widehat {A'CX})=\frac A2\ .$
Standard notation. $\boxed {\ (XY\ }$ - the ray without the point $X$ .

This post has been edited 1 time. Last edited by Virgil Nicula, May 29, 2006, 12:26 AM

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Megus

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Megus

#7 h May 28, 2006, 6:54 PM • 5 Y

Y by pog, jhu08, ImSh95, Adventure10, Mango247

Thank you, but I don't see how to apply this lemma

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treegoner

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treegoner

#8 h May 28, 2006, 8:23 PM • 4 Y

Y by pog, Adventure10, jhu08, ImSh95

Another application of spiral similarity.

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Virgil Nicula

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Virgil Nicula

#9 h May 29, 2006, 12:24 AM • 5 Y

Y by pog, Adventure10, jhu08, ImSh95, Mango247

You are a clever young because you are right, Megus ! See more above the my first reply where is the complete solution of this problem.

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darij grinberg

6555 posts

darij grinberg

#10 h May 29, 2006, 3:02 PM • 5 Y

Y by pog, Adventure10, jhu08, ImSh95, Mango247

BogG wrote:

Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$ . Let $H$ be the orthocenter of triangle $ABC$ , and let $M$ be the midpoint of the side $BC$ . Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$ , $H$ , $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$ .

First, I repost a solution given by Andre.L which was deleted for an unclear reason (to anyone who feels spoken to: according to Valentin, these things are not supposed to be deleted!).

andre.l wrote:

The common chord of the cirmcumcircles of $ABC$ and $ADE$ is their radical axis. Thus, it's perpendicular to the line that passes through their centers. Let $O$ and $O_1$ be the circumcenters of $ABC$ and $ADE$ , respectively. Then, $HM$ is perpendicular to the common chord of the circumcircles of $ABC$ and $ADE$ iff $HM$ is parallel to $OO_1$ . Let $Q=OO_1\cap AH$ . Since $OM\bot BC$ and $AH\bot BC$ , $OM\parallel AH$ . Thus, $OO_1\parallel HM \Leftrightarrow OQ \parallel HM \Leftrightarrow OMHQ$ is a parallelogram. Let $S$ be the midpoint of $OH$ . Then, $OMHQ$ is a parallelogram $\Leftrightarrow MS=SQ$ . But the circle with center $S$ which passes through $M$ is the nine-point circle of $ABC$ . Thus, it intersects $AH$ at it's midpoint. What we will prove is that $Q$ is the midpoint of $AH$ .
The point $O_1$ is on the perpendicular bisector of $DE$ , which is also the angle bisector of $BAC$ . Let $T$ be the intersection of $OH$ and the angle bisector of $BAC$ . Now, apply Menelaus' theorem to the line $OO_1Q$ and the triangle $HTA$ . We get that
$\frac{OH}{OT} \frac{O_1T}{O_1A} \frac{QA}{QH}=1$ .
Thus, $QA=AH \Leftrightarrow \frac{OH}{OT} \frac{O_1T}{O_1A} =1$ .
With some calculation, we can prove that the above relation is equivalent to $\cos A=2\cdot\cos^2\left(\frac{A}{2}\right)-1$ , which is true.

Let me add my solution.

Let B' and C' be the feet of the altitudes of triangle ABC from the vertices B and C, respectively. Keep in mind that these altitudes intersect at the orthocenter H of triangle ABC. Since < BB'C = 90° and < BC'C = 90°, the points B' and C' lie on the circle with diameter BC; the center of this circle is the midpoint M of the segment BC. So the circle with diameter BC is a circle with center M which passes through the points B and C and intersects the sides CA and AB again at the points B' and C', respectively. Hence, according to IMO 1985 problem 5, if R is the point of intersection of the circumcircles of triangles ABC and B'AC' (apart from A), then < MRA = 90°. Thus, the point M lies on the perpendicular to the line AR at the point R. On the other hand, since < AB'H = 90° and < AC'H = 90°, the points B' and C' lie on the circle with diameter AH. Consequently, the circumcircle of triangle B'AC' is the circle with diameter AH; therefore, since the point R lies on this circumcircle, it satisfies < ARH = 90°. Consequently, the point H lies on the perpendicular to the line AR at the point R.

Now as we know that both points M and H lie on the perpendicular to the line AR at the point R, we obtain $HM\perp AR$ . Now we will prove that the point R lies on the circumcircle of triangle ADE. Once this is shown, it will follow that the points A and R are the two common points of the circumcircles of triangles ABC and ADE; this will mean that the line AR is the common chord of these circumcircles, so that the relation $HM\perp AR$ will tell us that the line HM is perpendicular to the common chord of the circumcircles of triangles ABC and ADE. This will solve the problem.

So it only remains to show that the point R lies on the circumcircle of triangle ADE.

Since AE = AD, the triangle DAE is isosceles, so that its base angle is $\measuredangle ADE=\frac{180^{\circ}-\measuredangle DAE}{2}=90^{\circ}-\frac{\measuredangle DAE}{2}=90^{\circ}-\frac{A}{2}$ . In other words, $\measuredangle C^{\prime}DH=90^{\circ}-\frac{A}{2}$ . Similarly, $\measuredangle B^{\prime}EH=90^{\circ}-\frac{A}{2}$ . Thus, < C'DH = < B'EH. Together with < DC'H = < EB'H (both of these angles are 90°), this yields that the triangles C'DH and B'EH are similar; hence, $\frac{DC^{\prime}}{C^{\prime}H}=\frac{EB^{\prime}}{B^{\prime}H}$ . On the other hand, since < BC'H = < CB'H (both of these angles are 90°) and < BHC' = < CHB', the triangles BC'H and CB'H are similar, so that $\frac{C^{\prime}H}{BC^{\prime}}=\frac{B^{\prime}H}{CB^{\prime}}$ . Thus, $\frac{DC^{\prime}}{BC^{\prime}}=\frac{DC^{\prime}}{C^{\prime}H}\cdot\frac{C^{\prime}H}{BC^{\prime}}=\frac{EB^{\prime}}{B^{\prime}H}\cdot\frac{B^{\prime}H}{CB^{\prime}}=\frac{EB^{\prime}}{CB^{\prime}}$ .

Now, since the point R lies on the circumcircle of triangle B'AC', we have < RC'A = < RB'A, so that < RC'B = 180° - < RC'A = 180° - < RB'A = < RB'C. On the other hand, since the point R lies on the circumcircle of triangle ABC, we have < RBA = < RCA, what becomes < RBC' = < RCB'. From < RC'B = < RB'C and < RBC' = < RCB', it follows that the triangles RBC' and RCB' are similar. The points D and E are corresponding points in these two similar triangles (since they lie on the respective sides BC' and CB' and divide them in the same ratio $\frac{DC^{\prime}}{BC^{\prime}}=\frac{EB^{\prime}}{CB^{\prime}}$ ); since corresponding points in similar triangles form equal angles, we thus have < RDC' = < REB', or, equivalently, < RDA = < REA. This yields that the point R lies on the circumcircle of triangle ADE, and the problem is solved.

Darij

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abdurashidjon

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abdurashidjon

#11 h May 31, 2006, 3:32 AM • 5 Y

Y by pog, Adventure10, jhu08, ImSh95, Mango247

Virgil Nicula wrote:

Here is a syntetical proof of this nice and difficult problem.

Can someone explain how to get this ones

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silouan

3954 posts

silouan

#12 h May 31, 2006, 8:16 AM • 6 Y

Y by pog, Adventure10, jhu08, ImSh95, Mango247, ehuseyinyigit

abdurashidjon wrote:

Virgil Nicula wrote:

Here is a syntetical proof of this nice and difficult problem.

Can someone explain how to get this ones

Ok .We have $AH=2OM=\sqrt{4R^2-BC^2}=\sqrt{4r^2-4R^2(sinA)^2}=2RcosA$

By the low of sines $\frac{AP}{sin(B+\frac{A}{2})}=2R$ so $AP=2Rsin(90+\frac{B-C}{2})$ so $AP=2R\cos \frac{B-C}{2}$

And $MP=OP-OM=R(1-cosA)$

BTW , mr Nicula there exist a not metrical solution to this lemma ?

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dhernandez

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dhernandez

#13 h Jun 6, 2006, 2:45 AM • 8 Y

Y by mathisreal, myh2910, pog, Adventure10, jhu08, ImSh95, Mango247, and 1 other user

BogG wrote:

Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$ . Let $H$ be the orthocenter of triangle $ABC$ , and let $M$ be the midpoint of the side $BC$ . Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$ , $H$ , $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$ .

All the problems from the Swiss Imo Selection Team 2006

This is problem is difficult, not because it's lacking swift solutions but because it's so easy to go pursuing some typical ideas who do provide you with ways to solve it but I think, more complicated ones. Here is my proof

1) Name $O$ and $O'$ the circumcenter of $ABC$ and of the other triangle. Let $F$ be the point where $AO$ intersects again the circumcircle of $ABC$ . First thing to notice is that if $MH$ is to be perpendicular to $AK$ then it will intersect it precisely at $K$ . This is because $MH$ passes through $F$ and hence, if it's going to cut $AK$ perpendicularly it must do it in a point who belongs to the big circle, ergo $K$ .
1a) Notice that line $OO'$ cuts $AK$ perpendicularly at the midpoint. Hence it is parallel to line $FK$ . It suffices to prove therefore that $H$ is in line $FK$ because then $M$ is in line $FH$ and they are all together beautifully.

2) With all this in mind the problem is actually half done. Triangles $BDH$ and $CEH$ are similar for $\angle HCE = \angle HBD$ and $\angle HDB = \angle HEC$ . Hence
$\frac{BD}{CE} = \frac{DH}{HE}$
3) $\angle DKE = \angle BAC = \angle BKC \Rightarrow \angle BKD = \angle CKE$
4) $\angle KBD = \angle KCE$ hence, using 3, triangles $KBD$ and $KEC$ are similar hence $\frac{KD}{KE} = \frac{BD}{CE}$
5) With 3 and 4 $KH$ actually bisects $\angle DKE$ and we're set because if $G$ is the point where it touches again the small circle the $G$ and $A$ are diametrically opposite and hence $\angle HKA$ is right which means $H$ is in line $FK$ .

Daniel

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shobber

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shobber

#14 h Aug 31, 2006, 9:31 AM • 5 Y

Y by pog, jhu08, ImSh95, Adventure10, Mango247

Here is a proof with the use of trignometry. It is quite long so I think that I had better hide it.

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FelixD

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FelixD

#15 h Jun 27, 2009, 7:14 PM • 5 Y

Y by pog, jhu08, ImSh95, Adventure10, Mango247

Let $O$ be the circumcenter, $R$ be the radius and $k$ be the circumcircle of $\triangle ABC$ and let $r$ be the radius and $k'$ be the circumcircle of $\triangle ADE$ . Denote by $S$ the intersection of $AO$ with $k$ , by $N$ the intersection of the line $OM$ with $k$ and let $T$ be the intersection of the line $AN$ with $k'$ . Moreover, let $X$ be the intersection of $DE$ with $AT$ . We'll prove that $M$ , $T$ , $H$ are collinear and that $AO$ and $MH$ intersect in $k$ . Then it will follow that the point of intersection of the line $MH$ with $k$ and with $k'$ will be a common point, let's say $P$ and that the angle $\angle HPA = 90$ , as desired.
It's well-known that $AX$ and $OM$ intersect in $N$ . Moreover, $\angle HAX = \angle XAO$ . First, we'll prove that $S \in MH$ . Let $H_1$ and $H_2$ be the reflections of $H$ on $AB$ and $AC$ , respectively. We have $HC \Vert BS$ and thus $BH = BH_1 = SL$ and $HB \Vert CS$ and thus $CH = CH_2 = BS$ . Therefore, $HBSC$ is a parallelogramm and it follows that $HS$ intersects $BC$ at its middle, i.e. $M$ , as desired.
It's left to prove that $T \in MH$ , i.e. $T \in SM$ . Applying Menelaos' theorem to the triangle $\triangle NAO$ , it's remains to prove that $AN/NO \cdot OM/MN \cdot NT/TA = - 1$ or equivalently $MN/2OM = TN/TA = AN/AT - 1$ . It's well-known that $2OM = AH$ , thus $MN/2OM = R - OM/2OM = R/2OM - 1/2 = R/AH - 1/2$ . Handling $AN/AT$ is the more difficult part. We have $AX = AH \cdot \cos\angle HAX$ and $AX = AE \cdot \cos \alpha/2$ . Thus, $AE = AH \cdot \cos \angle HAR / \cos \alpha/2$ . From the equality of the area of triangle $\triangle ADE$ it follows that $DE = AE^2 \cdot \sin \alpha / AR \Leftrightarrow DE = AH \cdot \sin \alpha \cos \angle HAX / \cos^2 \alpha/2$ . But $AT = DE/\sin \alpha$ and therefore $AT = AH \cdot \cos \angle HAX / \cos^2 \alpha/2$ . Moreover, $AN = 2R \cdot \cos \angle XAO$ . Thus, $AN/AT = 2R \cdot \cos^2 \alpha/2 /AH$ . It's left to prove that $1/2 = R/AH \cdot (2\cos^2 \alpha/2 - 1)$ . We have $AH/2R = OM/R = \cos \alpha$ . Therefore it's left to prove that $\cos \alpha + 1 = \cos^2 \alpha/2$ , which follows from the addition theorems.

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Deadline

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Deadline

#124 h Mar 28, 2024, 7:00 AM

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Aiden-1089

350 posts

Aiden-1089

#125 h May 1, 2024, 6:17 AM • 1 Y

Y by Om245

Let $Q$ be the $A$ -Queue point, $X$ be the midpoint of $DE$ so $AX \perp DE$ , $A_1$ be the foot of the altitude from $A$ , and $T$ be the point on $BC$ such that $(A_1,T;B,C)=-1$ . It is well-known that $A,Q,T$ are collinear.
Let $G$ be the intersection between $AT$ and $DE$ , so $-1=(A_1,T;B,C) \stackrel{A}{=} (H,G;D,E)$ . Now $GD \cdot GE = GH \cdot GX = GA \cdot GQ$ , so $A,Q,D,E$ are concyclic.
Since $AQ \perp HM$ , we are done.

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shendrew7

818 posts

shendrew7

#126 h Jun 17, 2024, 2:46 AM

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Define $X$ , $Y$ as the feet of the altitudes from $B$ , $C$ . Firstly, we angle chase to find that $\triangle BHY \sim \triangle CHX$ and $HD$ , $HE$ bisect $\angle BHY$ , $\angle CHX$ , so we have the ratio
$\frac{DY}{DB} = \frac{EX}{EC} \implies \frac{\operatorname{pow}(D,(AH))}{\operatorname{pow}(D,(ABC))} = \frac{\operatorname{pow}(E,(AH))}{\operatorname{pow}(E,(ABC))}.$
Thus $(AH)$ , $(ADE)$ , $(ABC)$ are coaxial through Coaxiality Lemma, with the radical axis being $AQ$ , where $Q$ is the $A$ -queue point, which is perpendicular to $HM$ . $\blacksquare$

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N3bula

316 posts

N3bula

#127 h Oct 3, 2024, 11:28 AM

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It suffices to prove that $(AH)\cap (ABC)$ lies on $(ADE)$ and let that intersection be $K$ , this is because we have that $MH$ passes through $(AH)\cap(ABC)$ and is perpendicular
to their radical axis. Let $X$ and $Y$ be the feet of the altitudes from $B$ and $C$ . Clearly we have $(AHXY)$ . Note that there is a spiral similarity
sending $CX$ to $BY$ centred at $K$ . Proving $DE$ is the angle bisector of $\angle BHY$ suffices as this implies $\frac{EX}{XC}=\frac{EH}{HC}=\frac{DH}{DB}=\frac{DY}{YB}$ which
implies the spiral similarity centred at $K$ also sends $EX$ to $DY$ which suffices. $DE$ is the angle bisector as we have $\angle XHE=90-\angle AEH = \frac{\angle BAC}{2}$ , and we
have that $\angle XHC=90-\angle HCX=\angle ABC$ so this suffices.

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Eka01

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Eka01

#128 h Oct 12, 2024, 6:05 AM • 1 Y

Y by alexanderhamilton124

The problem condition is equivalent to $(ADE)$ Passing through the $A$ queue point of $\Delta ABC$ which we call $Q$ . Since $Q$ is the center of spiral similarity sending $XY$ to $BC$ , where $X$ and $Y$ are feet of $B$ and $C$ altitudes respectively, it follows that must also send $DE$ to $BC$ or $BD$ to $CE$ for the problem condition to be true, but that follows from some ratio chasing so we are done.

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AN1729

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AN1729

#129 h Oct 23, 2024, 12:33 PM

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Let $E$ and $F$ be the feet of altitudes from $B$ and $C$ respectively
$\triangle BHF \sim \triangle CHE$
$\angle FMH = \angle ENH \implies \frac{BM}{MF} = \frac{CN}{NE}$
Let $Q$ be the A-queue point
Consider the spiral similarity centered at $Q$ sending $\overline{BF}$ to $\overline{CE}$
Clearly, $M \longrightarrow N$
Thus, $Q$ is also center of spiral similarity sending $\overline{BM}$ to $\overline{CN}$
$\implies Q \in (AMN)$
It is now well known that $Q-H-D$ and $HD \perp AQ$
$\blacksquare$

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cj13609517288

1939 posts

cj13609517288

#130 h Nov 12, 2024, 9:04 PM

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Let $X$ and $Y$ be the feet of the altitudes from $B$ and $C$ , respectively.

Note that
$\angle YHD=90^{\circ}-\angle ADH=\frac12\angle A.$ Similarly, $\angle XHE=\angle YHD$ . Also, since $\angle YHB=\angle XHC$ , we also get $\angle DHB=\angle EHC$ . Since $\triangle YBH\sim\triangle XCH$ , we get that $\frac{YD}{DB}=\frac{XE}{EC}$ .

Let $P$ be the A-queue point; then it is on $(AXY)$ and $(ABC)$ . We want to show that it is also on $(ADE)$ , but since $D$ maps to $E$ in the same spiral similarity as $BY\to CX$ (by our ratio equivalence earlier), we get
$\angle DPE=\angle BPC=\angle BAC=\angle DAE,$ so indeed $P$ lies on $(ADE)$ .

Finally, since $AP\perp HM$ (since $\angle APH=90^{\circ}$ and the $A$ -antipode lies on $HM$ ), we are done. $\blacksquare$

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bin_sherlo

755 posts

bin_sherlo

#131 h Nov 24, 2024, 1:46 PM • 2 Y

Y by Maksat_B, tiny_brain123

Denote by $Q$ , $A-$ queue point which lies on $MH$ . If $H_b,H_c$ are the altitudes from $B,C$ then $\measuredangle BHE=\frac{\measuredangle A}{2}=\measuredangle EHH_C$ and similarily $HF$ bisects $\measuredangle H_BHC$ .
$\frac{Pow(E,(AHQ))}{Pow(E,(ABCQ))}=\frac{EH_C.EA}{EA.EB}=\frac{HH_C}{HB}=\frac{HH_B}{HC}=\frac{Pow(F,(AHQ))}{Pow(E,(ABCQ))}$ Now, coaxiality lemma finishes it as desired. $\blacksquare$

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ihategeo_1969

284 posts

ihategeo_1969

#132 h Jan 22, 2025, 8:42 PM

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See that due to the fact that the $A$ -queue point lies on the line $\overline{HMA'}$ where $A'$ is the $A$ -antipode, it suffices to prove that $(ABC)$ , $(ADE)$ , $(AH)$ are coaxial.

Let $Y$ and $Z$ be foot of altitudes of $B$ and $C$ . So due to the orz coaxiality lemma we just need to prove that $\frac{\text{Pow}(D,(AH))}{\text{Pow}(D,(ABC))}=\frac{\text{Pow}(E,(AH))}{\text{Pow}(E,(ABC))} \iff \frac{DZ}{DB}=\frac{EY}{EC}$ But see that this is trivial because $\triangle HZB \cup D \overset{-}{\sim} \triangle HYC \cup E$ and done.

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Nari_Tom

117 posts

Nari_Tom

#133 h Apr 19, 2025, 5:23 PM

Y by

Another easy approach is here.

Let $Y$ and $Z$ be the intersection of $B$ and $C$ altitudes with $(ABC)$ . Let $N$ be the midpoint of minor arc $BC$ . Then $D'=NZ \cap AB$ , $E'=NY \cap AC$ . By Pascal's theorem $D'-H-E'$ is collinear, and it's easy to see that $AD'=AE'$ . So these are the points in the original question. Let $G$ be the antipode of $A$ in $(ADE)$ . Let $P=DG \cap BH$ , $Q=EG \cap CH$ . Some angle chase gives $\triangle DBH \sim \triangle ECH$ , and $P$ and $Q$ are the corresponding points in this similarity. Thus $PQ \parallel BC$ $\implies$ $H-G-M$ collinear. Conclusion follows.

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alexanderchew

31 posts

alexanderchew

#134 h May 3, 2025, 4:07 AM

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after inversion this problem reduces to lengthbashing

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OronSH

1841 posts

OronSH

#135 h May 8, 2025, 6:38 PM

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Replace $D,E$ with $X,Y$ and let $\triangle DEF$ be the orthic triangle. Since $XY$ bisects $\angle BHF$ , it follows $BHFX\sim CHEY$ , so $\frac{BX}{FX}=\frac{CY}{EY}$ . Now if $Q$ is the queue point, by gliding principle it follows that $AQXY$ is cyclic. It is well known $AQ\perp HM$ , so we are done.

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Ilikeminecraft

734 posts

Ilikeminecraft

#136 h May 12, 2025, 11:24 PM

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Construct feet from $B$ to $AC$ and $C$ to $AB$ as $E_1, F_1.$
We have that $DE$ is perpendicular to the angle bisector of $BAC,$ and thus, $DE$ is the angle bisector of $F_1HB.$ However, $HF_1B\sim HE_1C,$ so $\frac{F_1D}{DB} = \frac{EE_1}{EC}.$ Simple spiral similarity combined with configuration facts on the Queue point finish.

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Giant_PT

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Giant_PT

#137 h May 13, 2025, 4:10 AM

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Let feet of altitude from $B$ to $AC$ be $X$ and from $C$ to $AB$ be $Y$ . Also, let the intersection of $(ABC)$ and $(ADE)$ other than $A$ be $P$ .
Claim: $\triangle BDH \sim \triangle CEH$
By angle chasing,
$\measuredangle DBH =\measuredangle YBX = \measuredangle YCX = \measuredangle HCE,$ $\measuredangle HDB =\measuredangle HDA = \measuredangle AEH = \measuredangle CEH,$ which finishes the claim. $\square$

Now, using claim 1 and the fact that $\triangle HBY \sim \triangle HCX$ , we can calculate that,
$\frac{BD}{BH}=\frac{CE}{CH}, \frac{BH}{BY}=\frac{CH}{CX} \implies \frac{BD}{BY} = \frac{CE}{CX}$ Since $P$ is the Miquel point of $BDEC$ , and spiral similarity is linear, we can see by using the previously calculated ratio that $P$ must also be Miquel point of $BYXC$ . Then, it is well-known by queue point configuration that $P$ , $H$ , and $M$ are collinear and thus $\measuredangle APM = \measuredangle AYH = 90^\circ$ $\square$ .

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This post has been edited 3 times. Last edited by Giant_PT, Jun 28, 2025, 10:01 PM

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alexanderchew

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alexanderchew

#138 h Jun 24, 2025, 11:48 AM

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There are two solutions that I know of, one which is Solution 1, and Solution 2

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