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Another orthocenter problem <3

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Source: Mexico National Olympiad 2019 P6

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#1 h Nov 12, 2019, 11:23 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle such that $\angle BAC = 45^{\circ}$ . Let $H,O$ be the orthocenter and circumcenter of $ABC$ , respectively. Let $\omega$ be the circumcircle of $ABC$ and $P$ the point on $\omega$ such that the circumcircle of $PBH$ is tangent to $BC$ . Let $X$ and $Y$ be the circumcenters of $PHB$ and $PHC$ respectively. Let $O_1,O_2$ be the circumcenters of $PXO$ and $PYO$ respectively. Prove that $O_1$ and $O_2$ lie on $AB$ and $AC$ , respectively.

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#2 h Nov 13, 2019, 4:00 AM • 1 Y

Y by Adventure10

hint

This post has been edited 1 time. Last edited by Al3jandro0000, Nov 13, 2019, 4:04 AM

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#3 h Nov 13, 2019, 10:20 AM • 8 Y

Y by ShinyDitto, Bassiskicking, Professor-Mom, ILOVEMYFAMILY, Adventure10, math_comb01, soryn, Hu_cosine

I have a very different and much shorter solution.

Let $A' = PH\cap\omega$ . Then using Reim's theorem with $\overline{PHA'}$ and $\overline{BBC}$ gives $CA'\parallel BH$ or $A'$ is the $A$ -antipode in $\omega$ . This means $\odot(PCH)$ is also tangent to $BC$ which means that it suffices to show that $O_1\in AB$ .

Now we use a common trick. We will reflect $O$ across $AB$ to get $O'$ and show that $P,X,O,O'$ are concyclic instead. Notice that $O'X$ and $OO'$ are perpendicular bisectors of $BH, BA$ respectively. Thus
$\angle XO'O = \angle ABH = 45^{\circ}$ On the other hand, we can easily chase $\angle XPO$ Since $OX$ is the perpendicular bisector of $BP$ , we get
$\angle XPO = \angle XBO = 90^{\circ} - \angle OBC = 45^{\circ}$ so we are done.

This post has been edited 1 time. Last edited by MarkBcc168, Nov 13, 2019, 10:21 AM

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#4 h Nov 15, 2019, 1:21 AM • 2 Y

Y by Adventure10, soryn

Notice that since $\angle BPC + \angle BHC = 180$ , we know from the condition of the problem that $H$ is the $P-$ HM point of $\triangle BPC.$ This means that $HP$ bisects $BC$ , and so $P \in HA'.$

We have $\angle PYC = 2 (180 - \angle PHC) = 2 \angle A'HC = 2 \angle PCB = \angle POB.$

Analogously $\angle PXB = \angle POC.$ This means that $PXBO$ and $POCY$ are similar kites. Hence, we've that $\angle PO_1O = 2 \angle PXO = 2 \angle POY = \angle POC.$

Similarly, $\angle PO_2O = \angle POB$ , so we get that $\triangle PO_1O \sim \triangle POC$ and $\triangle PO_2O \sim \triangle POB.$

We are now set up to begin the complex bash. WLOG let $B = 1, C = i, A = a$ for some complex number $a$ on the unit circle. If we let $E, F$ be the feet of $B, C$ onto $AC, AB$ respectively, then we know that $\triangle PEF \sim \triangle PCB.$ We have $F = \frac{a+i+1-ai}{2}$ and $E = \frac{a+i+1+ai}{2}$ . This means that $P = \frac{EB - CF}{E + B - F - C} = \frac{ai - 1 - a}{ai - 1 + i}$ .

Hence, we have from $\triangle PO_1O \sim \triangle POC$ that $O_2 = P + \frac{P^2}{1-P} = \frac{P}{1-P} = \frac{ai-1-a}{a+i}.$

To show that $A, O_2, C$ are collinear, we need to show that:

$\frac{a - \frac{ai-1-a}{a+i}}{a - i} \in \mathbb{R}.$
This simplifies to $\frac{a^2+a+1}{a^2+1}$ , and this is clearly equal to its complex conjugate, hence it's real.

Therefore, $O_2 \in AC$ . Analogously, $O_1 \in AB$ and we're done.

$\square$

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#6 h Nov 16, 2019, 8:51 PM • 3 Y

Y by Adventure10, Mango247, soryn

I am only going to prove $O_1 \in AB$ since the other case is anologous. Once you get $POO_1 \sim PCO$ , write: $\frac{PC}{CO} = \frac{PO}{OO_1} = \frac{AO}{OO_1}.$ Let $P'$ be the point where the parallel to $AB$ through $P$ cuts $\omega$ and let $A'$ be the point where $AH$ cuts $\omega$ . Notice that $HO \perp CP'$ , thus $HP'=HC$ . Furthermore $CHP' \sim COP$ . Thus we write: $\frac{PC}{CO} = \frac{P'C}{CH} = \frac{P'C}{CA'}.$ Finally $\angle AOO_1 = \angle P'CA' \implies AOO_1 \sim P'CA' \implies \angle O_1AO= \angle A'P'C = \angle BAO \implies O_1 \in AB$ .

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#7 h Nov 19, 2019, 6:53 AM • 2 Y

Y by Adventure10, Mango247

Bonus problem: In the same configuration, let $B'$ and $C'$ be the $B$ and $C$ -antipodes respectively. Show that the circumcenter of $AO_1O_2$ lies on $B'C'$ .

(This was actually included in the original proposal for this problem, but it was deemed too difficult for the National Olympiad)

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#8 h Nov 20, 2019, 12:01 AM • 1 Y

Y by Adventure10

juckter wrote:

Bonus problem: In the same configuration, let $B'$ and $C'$ be the $B$ and $C$ -antipodes respectively. Show that the circumcenter of $AO_1O_2$ lies on $B'C'$ .

(This was actually included in the original proposal for this problem, but it was deemed too difficult for the National Olympiad)

We can use MarkBcc168's trick. Reflect $A$ across $B'C'$ to get $K$ . We may easily check that $\angle O_1C'K = \angle O_2B'K$ . Also: $\frac{O_2B'}{O_1C'} = \frac{O_2P}{O_1P} = \frac{CP}{BP} = \frac{BA'}{CA'} = \frac{B'K}{C'K}.$ $\implies O_1C'K \sim O_2B'K \implies O_1KO_2 \sim C'KB' \implies \angle O_1KO_2=135^{\circ} \implies AO_1KO_2 \text{ cyclic}$ .

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#9 h Nov 20, 2019, 12:28 AM • 1 Y

Y by Adventure10

juckter wrote:

Bonus problem: In the same configuration, let $B'$ and $C'$ be the $B$ and $C$ -antipodes respectively. Show that the circumcenter of $AO_1O_2$ lies on $B'C'$ .

(This was actually included in the original proposal for this problem, but it was deemed too difficult for the National Olympiad)

We might also let $O_3$ be said circumcenter. Check that $O_3$ lies on the circumcircles of $PC'O_1$ and $PB'O_2$ . Furthermore $O_3$ lies on $B'C'$ .

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#10 h Jul 26, 2020, 3:35 AM • 3 Y

Y by v4913, ChCar, Rounak_iitr

Solution from Twitch Solves ISL:

We let $\overline{BK}$ and $\overline{CL}$ be the altitudes of $\triangle ABC$ . The circle with diameter $\overline{BC}$ will be denoted by $\gamma$ ; and we'll denote the center by $M$ .

Claim: The circle $(PHC)$ is also tangent to line $BC$ .
Proof. We are given $\measuredangle PBC = \measuredangle PHB$ . Since $\measuredangle BHC = -\measuredangle BAC = -\measuredangle BPC$ , it follows $\measuredangle PCB = \measuredangle PHC$ too. $\blacksquare$

Claim: Points $P$ , $H$ , $M$ are collinear. Actually, $P$ is the inverse of $H$ with respect to $\gamma$ .
Proof. Line $PH$ bisects $\overline{BC}$ by radical axis on $(BPH)$ and $(CPH)$ . Also, $MH \cdot MP = MB^2 = MC^2$ . $\blacksquare$

$[asy]size(10cm); pair B = dir(180); pair C = dir(0); pair K = dir(65); pair L = dir(90)*K; filldraw(unitcircle, invisible, blue); filldraw(B--C--K--L--cycle, invisible, blue); pair A = extension(B, L, C, K); pair H = extension(B, K, C, L); draw(B--K, blue); draw(C--L, blue); pair O = circumcenter(A, B, C); pair P = 1/conj(H); pair M = origin; filldraw(circumcircle(P, B, H), invisible, orange); pair X = circumcenter(P, H, B); pair O_1 = circumcenter(P, X, O); draw(L--A--K, dotted+deepcyan); draw(circumcircle(X, O, L), dashed+deepgreen); draw(P--M, blue); draw(CP(O_1, P), dotted+grey); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$A$", A, dir(A)); dot("$H$", H, dir(110)); dot("$O$", O, dir(60)); dot("$P$", P, dir(P)); dot("$M$", M, dir(-90)); dot("$X$", X, dir(X)); dot("$O_1$", O_1, dir(O_1)); /* TSQ Source: B = dir 180 R225 C = dir 0 R315 K = dir 65 L = dir(90)*K unitcircle 0.1 lightcyan / blue B--C--K--L--cycle 0.1 lightcyan / blue A = extension B L C K H = extension B K C L R110 B--K blue C--L blue O = circumcenter A B C R60 P = 1/conj(H) M = origin R-90 circumcircle P B H 0.1 yellow / orange X = circumcenter P H B O_1 = circumcenter P X O L--A--K dotted deepcyan circumcircle X O L dashed deepgreen P--M blue CP O_1 P dotted grey */ [/asy]$
We now actually use the condition that $\measuredangle BAC = 45^{\circ}$ , which is equivalent to $\measuredangle BHC = 135^{\circ}$ and $\measuredangle BOC = 90^{\circ}$ . This means that $O$ is the arc midpoint of $(BC)$ .

Claim: $LOKH$ is a parallelogram.
Proof. Since arcs $KL$ and $OB$ of $\gamma$ measure $90^{\circ}$ , the arcs $BL$ and $OK$ are equal, hence $\overline{LO} \parallel \overline{BK}$ . Similarly $\overline{KO} \parallel \overline{CL}$ . $\blacksquare$

Claim: We have $\measuredangle XPO = 45^{\circ}$ .
Proof. Since $\triangle HLK \sim \triangle HBC$ , and $\overline{HO}$ bisects $\overline{LK}$ by previous claim, it follows $\measuredangle BHM = \measuredangle OHL.$ Now $\begin{align*} \measuredangle XPO &= \measuredangle XPH + \measuredangle MPO = (90^{\circ} - \measuredangle HBP) + \measuredangle HOM \\ &= 90^{\circ} - \measuredangle MBP + \measuredangle MBH + \measuredangle HOM \\ &= 90^{\circ} - \measuredangle BHM + \measuredangle MBH + \measuredangle HOM \\ &= 90^{\circ} - \measuredangle OHL + \measuredangle MBH + \measuredangle HOM \\ &= 90^{\circ} + \measuredangle(LH, HO) + \measuredangle(MB, BH) + \measuredangle(HO,MO) \\ &= 90^{\circ} + \measuredangle(LH, BH) + \measuredangle(MB, MO) = 45^{\circ}. \end{align*}$ $\blacksquare$

Claim: We have $X$ , $O_1$ , $O$ , $L$ , $H$ are concyclic, in the circle with diameter $\overline{XO}$ .
Proof. First, since $\triangle LBH$ is a $45^{\circ}$ - $45^{\circ}$ - $90^{\circ}$ triangle and $XB=XH$ , it follows that $\overline{XL}$ is the perpendicular bisector of $\overline{BH}$ . Hence, $\measuredangle XLO = 90^{\circ}$ .
On the other hand, since $\measuredangle XPO = 45^{\circ}$ , we have $\measuredangle XO_1O = 90^{\circ}$ . This implies concyclic. $\blacksquare$

Finally, $\measuredangle BLO = 135^{\circ}$ and $\measuredangle OLO_1 = \measuredangle OXO_1 = 90^{\circ} - \measuredangle XPO = 45^{\circ}$ this implies $O_1$ lies on line $BL$ , ergo lies on line $AB$ .

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L567

#12 h Sep 5, 2021, 5:06 PM • 1 Y

Y by geometry6

Solved with geometry6, mueller.25

Let $P'$ be the $H$ humpty point in $\triangle HBC$ . So, $\angle BP'C = \angle BP'H + \angle CP'H = \angle HBC + \angle HCB = \angle BAC$ and so $P' \in (ABC)$ and since $(P'BH)$ is tangent to $BC$ , we have $P' = P$ , which gives $(PHC)$ is tangent to $BC$ too. So, it suffices to show $O_1 \in AB$ .

Let $BX \cap (ABC) = Q$ . Since $\angle XQP = \angle BQP = \frac{\angle BOP}{2} = \angle XOP$ , we have $PQOX$ is cyclic. So, $\angle XPO = \angle XQO = \angle BQO = 90 - \angle OBC = 45^\circ$ .

Define $O'$ to be the reflection of $O$ across line $AB$ . It suffices to show that $O' \in (PXO)$ , or that $\angle XO'O = 45^\circ$ . But note that $O'$ is the center of $(AHB)$ , so $\angle XO'O = \angle HBA = 45^\circ$ , and so we are done. $\blacksquare$

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Eyed

#13 h Nov 13, 2022, 2:02 AM • 1 Y

Y by hyc2021

Complex cuz

Let $A', C'$ be the antipode of $A, C,$ and $M$ the midpoint of $BC$ . Note that
$\angle BPA' = 90 - \angle C = \angle HBC = \angle BPH$ so $P, H, A'$ are collinear. It is well known that $M$ lies on this line. Next, note that $OX$ perpendicularly bisects $PB$ so
$\angle PXO = 90 + \angle PHB = 90 + \overarc{PB} + 45 = 135 + \overarc{PB} = 180 - (45 + \overarc{PB}) = 180 - \angle PC'O$ so $(PXOC')$ concyclic. Now, we set complex numbers with the circumcenter as the origin, $C = -1, B = i, C' = 1, A' = a$ . We have $M = \frac{i-1}{2}, A = -a$ , and
$ap\overline{m} + m = a + p \implies p\left(a\frac{1+i}{2} + 1\right) = \frac{i-1}{2} -a \implies p = \frac{i-1-2a}{a(i+1)+2}$ $O_1$ is the circumcenter of $OPC'$ . Since $C' = 1$ , then $o_1 = \frac{p}{p+1} = \frac{i-1-2a}{i+ai + 1 - a}$ . Now, we want to show that $O_1\in AB$ , or $-ai\overline{o_1} + o_1 = -a + i$ . However, this resolves to
$-ai\overline{o_1} + o_1 = \frac{-ai(-i-1-\frac{2}{a})(-a)}{(-a)(-i - \frac{1}{a}i + 1 - \frac{1}{a})} + \frac{i-1-2a}{i+ai+1-a} = \frac{ai(-ai-a-2) + i-1-2a}{i+ai + 1 -a} = i-a.$ We conclude $O_1\in AB$ . Similarly, because $\angle HPC = \angle A'PC = 90 - \angle B = \angle HCB$ , we have $(PHC)$ is tangent to $BC$ as well, so by symmetry, $O_2\in AC$ .

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gvole

#14 h Apr 13, 2023, 11:14 AM

Y by

Like everyone else, we can show $P$ is the $A$ - Queue point, also $(PHC)$ tangent to $BC$ . We just want to show $O_1\in AB$ . Notice that the antipode $C'$ of $C$ is on $(POX)$ due to some angles. We will now use moving points.

Animate $A$ on $(BAC)$ projectively. Let $\ell$ be the bisector of $C'O$ .

Clearly $A\to AB\to \ell \cap AB$ is projective.

Also $A\to A' \to A'M \cap (ABC)=P\to P'\to P'O\cap \ell$ is projective, where $A'$ is the antipode of $A$ , $M$ is the midpoint of $BC$ , and $P'$ is the midpoint of $C'P$ .

We want to show those two bisectors intersect on $AB$ , so it's enough to check for 3 cases of $A$ . It's easy to verify what happens when $A$ is the antipode of $B$ or $C$ , and also when $A$ is the midpoint of major arc $BC$ .

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CT17

#15 h Jan 29, 2024, 2:36 AM

Y by

Let $A'$ be the $A-$ antipode. Observe that

$\measuredangle BPH = \measuredangle CBH = \measuredangle BAA' = \measuredangle BPA'$
so $P$ is actually the $A-$ orthic Miquel point. Now let $O'$ be the reflection of $O$ over $AB$ . As $OX$ is the perpendicular bisector of $BH$ , so we have

$\measuredangle XO'O = \measuredangle HBX = 90^\circ - \measuredangle BAC$
and

$\measuredangle XPO = \measuredangle BPO + \measuredangle XPB = (90^\circ + \measuredangle BA'P) - (90^\circ + \measuredangle PHB = \measuredangle BA'H + \measuredangle A'HB = \measuredangle A'BH = \measuredangle BAC$
so from $\angle BAC = 45^\circ$ it follows follows that $O'$ lies on $(PXO)$ . Since $AB$ is the perpendicular bisector of $OO'$ , we are done.

This post has been edited 1 time. Last edited by CT17, Jan 29, 2024, 2:37 AM

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#16 h Aug 12, 2024, 7:00 PM

Y by

Rename $P$ to $Q$ since it is the queue point. Let $O'$ be the reflection of $O$ over $AB$ and $F$ the foot from $C$ to $AB$ . Let $H$ be the orthocenter of $ABC$ . Let $AB$ hit $(BHQ)$ again at $J$ .

Claim: $FXJH$ is cyclic
Proof. First, note that $\angle JHF = 90^{\circ}$ . Then, $\angle JXH = 2\angle JBH=2\angle ABH = 90^{\circ}$ as desired. $\blacksquare$
Note by reflection that $\angle XQO=\angle XBO=45 ^{\circ}$ , so it suffices to show $\angle XO'O=45^{\circ}$ . But by the claim, as $\angle XFB=135 ^{\circ}$ and $\angle BFO=180^{\circ} - \angle OCB = 45 ^{\circ}$ we get that $F,X,O'$ collinear, so $\angle XO'O=\angle FO'O=\angle FOO'=90^{\circ} - \angle FA = 45 ^{\circ}$ as desired.

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Ywgh1

#17 h Aug 15, 2024, 7:25 PM

Y by

Mexico 2019/4
Label the points as the following.
We divide the problem into four claims.

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Claim 1: $P$ is the $A$ -queue point of $\triangle ABC$ .
Proof: Easy angle chase gives us that $BC$ is tangent to $(PHC)$ , then by PoP we have that $MC^2=MB^2=MH \cdot MP$ . $\blacksquare$

Claim 2: $HOH_1H_2$ is a parallelogram.
Proof: $\angle BMH_1=\angle OMH_2$ hence $\angle OH_1H_2=\angle BH_1H_2$ hence $OH_2 \| HH_1$ . And similarly $OH_1 \| HH_2$ . $\blacksquare$

Claim 3: $\angle XO_1O=90$ .
Proof: Long angle chase. $\blacksquare$

Finally we how our last claim.
Claim 4: $OH_2O_1X$ is cyclic.
Proof: By our previous claim $\angle XO_1O=90$ , and we also have that $\angle XH_2O=90$ since $XH_2$ is perpendicular to $BH$ , hence it’s cyclic. $\blacksquare$

Now since $\angle O_1H_2O= \angle H_2BH=45$ and $\angle O_1XO=45$ , hence we have that $O_1$ must lie on $AB$ . So we are done.

This post has been edited 5 times. Last edited by Ywgh1, Aug 15, 2024, 7:52 PM

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#18 h Oct 8, 2024, 10:02 AM • 2 Y

Y by cursed_tangent1434, idkk

truly rizztastic problem
solution

This post has been edited 1 time. Last edited by InterLoop, Oct 8, 2024, 10:03 AM

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#19 h Oct 9, 2024, 3:46 PM • 3 Y

Y by InterLoop, idkk, stillwater_25

Solved with stillwater_45. Very interesting problem. After I realized that $P$ with the $A-$ Queue Point of $\triangle ABC$ I thought this could turn into a orthocenter-configuration problem. Guess my wish was not to be granted.

Denote by $B_1$ and $C_1$ the feet of the altitudes from $B$ and $C$ . We can start off by noting that $O$ lies on $(BC)$ as well since we have $\measuredangle COB = 2\measuredangle BAC = \frac{\pi}{2}$ . Now, we prove the following claim, invoking some symmetry into the problem.

Claim : Circle $(PHC)$ is also tangent to $\overline{BC}$ and in particular, $P$ is the $A-$ Queue Point of $\triangle ABC$ .
Proof : Let $P'$ denote the intersection of the circles tangent to $\overline{BC}$ passing through points $B$ and $H$ and $C$ and $H$ respectively. Note that,
$\measuredangle CPB = \measuredangle CPH + \measuredangle HPB = \measuredangle BCH + \measuredangle HBC = \frac{\pi}{4}$ which implies that $P'$ lies on $(ABC)$ and indeed $P'\equiv P$ . Thus, the first part of the claim is proved. For the second part simply note that, since the radical axis $\overline{PH}$ of circles $(PHB)$ and $(PHC)$ must bisect the common tangent $\overline{BC}$ , $\overline{PH}$ passes through the midpoint of side $BC$ . It is well known that this implies that $P$ is the $A-$ Queue Point of $\triangle ABC$ .

Also note that,
$\measuredangle OPX = \measuredangle BPX + \measuredangle OPB = \frac{\pi}{2}+\measuredangle PHB + \frac{\pi}{2} + \measuredangle BCP = \measuredangle PHC_1 + \measuredangle C_1HB + \measuredangle BAP = \frac{\pi}{4}$ A similar calculation yields that $\measuredangle YPO = \frac{\pi}{4}$ as well.

With the symmetry now established and the above observations, we can prove the following results.

Claim : Quadrilaterals $XO_1OC_1$ and $YO_2OB_1$ are cyclic.
Proof : We prove the first one, the other follows similarly due to symmetry. First note that,
$\measuredangle OO_1X = 2\measuredangle OPX = \frac{\pi}{2}$ so it suffices to show that $\measuredangle XC_1O=\frac{\pi}{2}$ as well, which we shall do as follows.

Note that $C_1$ lies on the perpendicular bisector of segment $BH$ (since $\triangle BC_1H$ is a 45-45-90 right triangle). Also, $X$ lies on this perpendicular bisector as it is the center of $(BHP)$ . Thus, $\overline{XC_1}$ is the perpendicular bisector of segment $BH$ implying that $XC_1 \perp BH$ . Further note that, $O$ and $C_1$ (due the 45-45-90 right triangle $\triangle AC_1B$ ) lie on the perpendicular bisector of $AC$ . Thus, $OC_1 \perp AC \perp BH$ , implying that $OC_1 \parallel BH$ . Thus, $XC_1 \perp OC_1$ and indeed $\measuredangle XC_1O = \frac{\pi}{2}$ as desired.

Now we are essentially done since,
$\measuredangle OC_1O_1 = \measuredangle OXO_1 = \frac{\pi}{4} = \measuredangle OC_1A$ which implies that $O_1$ lies on $AB$ . A similar argument shows that $O_2$ also lies on $AC$ which finishes the proof of the problem.

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#20 h Dec 29, 2024, 3:01 AM

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Nice! Let $A'$ be the $A$ -antipode and $C'$ be the $C$ -antipode.

Claim: $P$ is the orthocenter Miquel point, i.e. it lies on $\overline{HA'}$ .

Proof: The actual Miquel point $P'$ satisfies $\measuredangle BP'A' = \measuredangle BAA' = \measuredangle CBH$ , i.e. $\overline{CB}$ is tangent to $(P'BH)$ . So $P = P'$ . $\blacksquare$

Claim: $\overline{BO}$ and $\overline{CP}$ intersect at the intersection point $Q$ of $(PXO)$ and $(PBH)$ .

Proof: $Q$ lies on $\overline{BO}$ as $\angle PQO + \angle PQB = 180^\circ - \frac 12 \angle PXB + \angle PQB = 180^\circ - \angle PQB + \angle PQB = 180^\circ.$ $Q$ lies on $\overline{CP}$ because $\measuredangle QPH = \measuredangle OBH = \angle C - 45^\circ = \measuredangle A'PC. \ \blacksquare$ Now note that $C'$ lies on $(POQX)$ because $CO \cdot CC' = 2BC^2$ as $\angle BOC = 90^\circ$ . In particular, $\angle COQ = 90^\circ$ , so it follows that $O_1$ is the midpoint of $\overline{C'Q}$ .

To finish, let $N$ be the midpoint of $\overline{AB}$ . Note that $\overline{QH} \parallel \overline{AB}$ by the same angle chase as before, so a homothety with ratio $\frac 12$ at $C'$ takes $\overline{QH}$ to $\overline{O_1N}$ . It follows that $O_1$ lies on $\overline{AB}$ , and the other part follows similarly.

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