Inversely Similiar Triangles

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EulerMacaroni

851 posts

EulerMacaroni

#1 h Apr 20, 2016, 9:30 PM • 7 Y

Y by Davi-8191, megarnie, Quidditch, HWenslawski, mathlearner2357, Adventure10, Mango247

Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\overleftrightarrow{BC}$ , and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$ , respectively.

Given that $AH^2=2\cdot AO^2,$ prove that the points $O,P,$ and $Q$ are collinear.

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djmathman

7938 posts

djmathman

#2 h Apr 20, 2016, 9:30 PM • 9 Y

Y by Generic_Username, sualehasif996, claserken, Ultroid999OCPN, Jc426, megarnie, Danielzh, Adventure10, Mango247

Pretty easy for a JMO5 in my opinion (and certainly easier than JMO1); it took me about 20 minutes to solve during DiffEq today.

Solution

This post has been edited 2 times. Last edited by djmathman, Apr 20, 2016, 9:35 PM

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EulerMacaroni

851 posts

EulerMacaroni

#3 h Apr 20, 2016, 9:31 PM • 7 Y

Y by anantmudgal09, amar_04, myh2910, megarnie, two_steps, Adventure10, Mango247

Anyone else thought this was much easier than JMO/1?

It is well-known that $AH\cdot 2AO=AB\cdot AC$ (just use similar triangles or $\sqrt{bc}$ inversion). Then by Power of a Point,
$AP\cdot AB=AH^2=AQ\cdot AC$ Consider the transformation $X\mapsto \Psi(X)$ which dilates $X$ from $A$ by a factor of $\dfrac{AB}{AQ}=\dfrac{AC}{AP}$ and reflects about the $A$ -angle bisector. Then $\Psi(O)$ clearly lies on $AH$ , and its distance from $A$ is $AO\cdot\frac{AB}{AQ}=AO\cdot\frac{AB}{\frac{AH^2}{AC}}=AO\cdot\frac{AB\cdot AC}{AH^2}=\frac{AO\cdot AH\cdot 2AO}{AH^2}=\frac{2AO^2}{AH}=AH$ so $\Psi(O)=H$ , hence we conclude that $O,P,Q$ are collinear, as desired.

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WL0410

408 posts

WL0410

#4 h Apr 20, 2016, 9:31 PM • 2 Y

Y by megarnie, Adventure10

Length bashing with trig to show PO+OQ=PQ!

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AMN300

563 posts

AMN300

#5 h Apr 20, 2016, 9:32 PM • 1 Y

Y by Adventure10

coordinate bashed gg

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Wave-Particle

3690 posts

Wave-Particle

#6 h Apr 20, 2016, 9:32 PM • 1 Y

Y by Adventure10

Complex bash also worked on this.

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Benq

3396 posts

Benq

#7 h Apr 20, 2016, 9:32 PM • 1 Y

Y by Adventure10

Barycentric bash also worked on this.

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mathwizard888

1635 posts

mathwizard888

#8 h Apr 20, 2016, 9:32 PM • 1 Y

Y by Adventure10

Coordinate bashed with $A=(0,a)$ , $B=(b,0)$ , $C=(c,0)$ , $H=(0,0)$ . It didn't turn out that bad, though I did get some messy expressions.

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zephyrcrush78

389 posts

zephyrcrush78

#9 h Apr 20, 2016, 9:33 PM • 2 Y

Y by Adventure10, Mango247

I bashed the area of triangle $\triangle PQO$ , which worked out pretty decently I guess.

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mathmaster2012

636 posts

mathmaster2012

#10 h Apr 20, 2016, 9:34 PM • 2 Y

Y by Adventure10, Mango247

Trig should show that O'A =R, where O' is intersection of PQ and reflection of AH over angle bisector. Is this sufficient?

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room456

146 posts

room456

#11 h Apr 20, 2016, 9:37 PM • 2 Y

Y by Adventure10, Mango247

definitely easier than jmo 1

first wlog R=1, then bc = 2sqrt2, then law of sines and law of cosines on AOP and AOQ to get that angles AOP and AOQ are right.

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mathwizard888

1635 posts

mathwizard888

#12 h Apr 20, 2016, 9:38 PM • 2 Y

Y by Adventure10, Mango247

Since the reflection of AH over the angle bisector is AO, this is sufficient, though you should add what I just said to the solution.

oops sniped

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bestwillcui1

2735 posts

bestwillcui1

#13 h Apr 20, 2016, 9:40 PM • 1 Y

Y by Adventure10

mathwizard888 wrote:

Coordinate bashed with $A=(0,a)$ , $B=(b,0)$ , $C=(c,0)$ , $H=(0,0)$ . It didn't turn out that bad, though I did get some messy expressions.

yes. mine was 6 pages , two exceptions when b and c are negatives or o and p are vertical

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FlakeLCR

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FlakeLCR

#14 h Apr 20, 2016, 9:42 PM • 3 Y

Y by seyyed_khandan, Adventure10, Mango247

Did anyone else prove that $P$ and $Q$ were on the diameter opposite of $A$ (immediately implying the solution)?

This post has been edited 1 time. Last edited by FlakeLCR, Apr 20, 2016, 9:43 PM
Reason: asdf

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mathwizard888

1635 posts

mathwizard888

#15 h Apr 20, 2016, 9:44 PM • 1 Y

Y by Adventure10

bestwillcui1 wrote:

mathwizard888 wrote:

Coordinate bashed with $A=(0,a)$ , $B=(b,0)$ , $C=(c,0)$ , $H=(0,0)$ . It didn't turn out that bad, though I did get some messy expressions.

yes. mine was 6 pages , two exceptions when b and c are negatives or o and p are vertical

Wait dealing with OP vertical is easy, and you know exactly one of b and c is negative because ABC is acute.

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MathLuis

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MathLuis

#104 h Nov 13, 2022, 2:30 PM

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Basicaly by the 90º angles we have that $(BPH),(CQH)$ are tangent at $H$ and the tangent line is $AH$ so by PoP $AP \cdot AB=AQ \cdot AC=AH^2=2AO^2=AO \cdot AA'$ where $A'$ is the $A$ -antipode in $(ABC)$ , now by inversion with center $A$ and radius $AH$ we have that $P \to B$ , $C \to Q$ , $O \to A'$ so $BPOA',CQOA'$ are cyclic but note that $(ABC) \to PQ$ so $O$ lies in $PQ$ , thus we are done

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HamstPan38825

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HamstPan38825

#105 h Dec 23, 2022, 8:16 PM

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The desired condition can be written as $[AOP] + [AOQ ] = [APQ]$ , or
$\begin{align*} R(AP \cos C + AQ \cos B) &= AP \cdot AQ \cdot \sin A \\ \sin^2 B\sin C \cos C + \sin^2 C \sin B \cos B &= 2\sin A \sin^3 B \sin^3 C \\ \sin^2 B \sin^2 C = \frac 12 \end{align*}$ as $AP = AH \sin B = 2R \sin^2 B \sin C$ . But $AH^2 = 2AO^2$ is obviously equivalent to this as well, so we are done.

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samrocksnature

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samrocksnature

#106 h Jan 5, 2023, 7:27 AM

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We can equivalently prove $[AOP] + [AOQ ] = [APQ],$ and using the sine area formula gives $R \cdot AQ \cdot \sin \angle BAO + R\cdot AP \cdot \sin \angle CAO = AQ \cdot AP \cdot \sin \angle A.$ Since $O$ and $H$ are isogonal conjugates, we note that $\angle BAO = \angle HAC = 90^\circ - \angle C$ and similarly $\angle CAO = \angle BAH =90^\circ - \angle B.$ Since $\sin (90^\circ-x)=\cos x,$ we simplify our equation to $R \cdot AQ \cdot \cos \angle C + R\cdot AP \cdot \cos \angle B = AQ \cdot AP \cdot \sin \angle A.$ In right triangle $AHP,$ observe that $\angle AHP=90^\circ - \angle HAP = \angle C,$ so $\sin \angle C = \tfrac{AP}{AH},$ but right triangle $AHC$ gives $\sin \angle C =\tfrac{AH}{b},$ hence equating gives $AP=\tfrac{AH^2}{b}=\tfrac{2R^2}{b}$ by the condition. Symmetrically, $AQ=\tfrac{2R^2}{c}.$ Substituting back in, $R \cdot \frac{2R^2}{c} \cdot \cos \angle C + R\cdot \frac{2R^2}{b} \cdot \cos \angle B = \frac{2R^2}{c} \cdot \frac{2R^2}{b} \cdot \sin \angle A$ which directly simplifies to $b \cos\angle C+c\cos \angle B=2R \sin \angle A=a,$ where the last step follows by the Extended Law of Sines. Now, expressing the cosines of the angles in terms of side lengths using the Law of Cosines and substituting, it remains to show $a=b \cdot \frac{a^2+b^2-c^2}{2ab}+c \cdot \frac{a^2+c^2-b^2}{2ac},$ which is trivial upon simplification.

Remarks: I needed a hint for the area re-interpretation, and the rest was blind trigging

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GrantStar

821 posts

GrantStar

#107 h Jan 22, 2023, 4:24 PM

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Invert around $A$ with radius $AH$ . Notice that line $PQ$ swaps with $(ABC)$ since $AP \cdot AB=AH^2$ by similar triangles. Let the inverse of $O$ be $O^*$ . From the condition, $AO\cdot AO^*=AH^2=2AO^2,$ so $O^*$ is the reflection of $A$ wrt $O,$ the antipode of $A$ . However, $O^*$ clearly lies on $(ABC),$ which means that $O$ lies on $PQ$ as desired.

This post has been edited 1 time. Last edited by GrantStar, Jan 22, 2023, 4:24 PM

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brainfertilzer

1831 posts

brainfertilzer

#108 h Mar 18, 2023, 5:42 PM

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trigbash ftw

Letting $R$ be the circumradius, the given length condition is $AH = \sqrt{2}R$ . Let $\angle HAB = x$ and $\angle HAC= y$ so that $\angle ABC = 90^\circ- x$ and $\angle ACB = 90^\circ - y$ . Additionally, note that $\angle OAP = 90^\circ - \angle ACB = y$ and similarly $\angle OAQ = x$ .

Now, to prove that $P, O, Q$ are collinear, it suffices to show that $[APQ] = [AOP] + [AOQ]$ . First, we will compute $[APQ]$ . Note that $AP = \sqrt{2}R\cos x$ and $AQ = \sqrt{2}R\cos y$ , so
$[APQ] = \frac{1}{2}(\sqrt{2}R\cos x)(\sqrt{2}R\cos y)\sin (x + y) = R^2\cos x\cos y\sin (x + y).$ Now, noting that $AO = R$ , we can also compute
$[AOP] = \frac{1}{2}AP\cdot AO\sin \angle OAP = \frac{\sqrt{2}}{2}R^2\cos x\sin y,$ and similarly $[AOQ] = \tfrac{\sqrt{2}}{2}R^2\cos y\sin x$ . It follows that $[AOP] + [AOQ] = \frac{\sqrt{2}}{2}R^2\sin (x + y)$ , so to show this equals $[APQ]$ , it suffices to show $\cos x\cos y = \tfrac{\sqrt{2}}{2}$ . Note that
$[ABC] = \frac{1}{2}AH\cdot BC= 2R^2\sin A\sin B\sin C,$ $\frac{1}{2}\sqrt{2}R\cdot 2R\sin A = 2R^2\sin A\sin B\sin C\implies \sin B\sin C = \frac{\sqrt{2}}{2}.$ However, $\sin B\sin C = \sin (90^\circ - x)\sin (90^\circ - y) = \cos x\cos y$ , so we have $\cos x \cos y = \tfrac{\sqrt{2}}{2}$ and we are done.

comment

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Spectator

657 posts

Spectator

#109 h Apr 14, 2023, 3:15 PM

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Let $O_{1}$ be the intersection of line $AO$ and $PQ$ . Note that $\angle{APQ} \cong \angle{AHQ} \cong \angle{ACB}$ because $APHQ$ is cyclic. Thus, $\triangle{APQ} \sim \triangle{ABC}$ .

Also note that $\angle{PAO_{1}} = 90-\angle{C}$ because of circumcircle properties. Thus, $\angle{PO_{1}A} = \angle{90}$ . Thus, $AO_{1}$ is the altitude.
$\frac{AQ}{AB} = \frac{AO_{1}}{AH}$
Note,
$AQ = AH\sin{C}$ and
$\frac{AB}{\sin{C}} = 2AO$ Thus,
$\begin{align*} AO_{1} &= \frac{AQ\cdot AH}{AB} \\ &= \frac{AH^2\sin{C}}{AB} \\ &= \frac{2AO^2}{2AO} \\ &= AO \end{align*}$ Thus, $O = O_{1}$ proving that $P$ , $O$ and $Q$ are collinear.

remarks

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huashiliao2020

1292 posts

huashiliao2020

#110 h Aug 1, 2023, 4:10 AM

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Let A' be the diametrically opposite point of A. Then $AO*AD=AH^2=AQ*AC=AP*AB,$ which readily implies the cyclic quads $BPOA',CQOA'\implies POA'+QOA'=180-PBA'+180-QCA'=180,$ as desired. $\blacksquare$

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Infinity_Integral

306 posts

Infinity_Integral

#111 h Sep 7, 2023, 1:40 AM

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Set circumcircle of $\triangle ABC$ as the unit circle and perform a Complex Numbers Coordinate bash. It works.

Full proof here:
https://infinityintegral.substack.com/p/usajmo-2016-contest-review

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joshualiu315

2534 posts

joshualiu315

#112 h Nov 19, 2023, 7:49 AM

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Let $AO = R$ . We have

$\begin{align*} OP^2 &= \left(AP-\frac{AB}{2} \right)^2 + \left(R^2 - \left(\frac{AB}{2} \right)^2 \right) \\ &= AP^2-AB \cdot AP +\frac{AB^2}{4} +R^2 - \frac{AB^2}{4} \\ & = AP^2- AH^2 +R^2 \\ &= AP^2-R^2. \end{align*}$
Hence, $\angle AOP = 90^\circ$ . We can also find that $\angle AOQ = 90^\circ$ in a similar fashion so we are done. $\square$

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bachkieu

137 posts

bachkieu

#113 h Feb 19, 2024, 3:31 PM

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Anyone try inversion about A with radius AH

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Sleepy_Head

566 posts

Sleepy_Head

#114 h Jun 18, 2024, 6:14 PM

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We proceed by complex with $O$ as the origin, and $A$ , $B$ , and $C$ lying on the unit circle.
By complex foot, we have
$h=\frac{1}{2}\left(a+b+c-\frac{bc}{a}\right).$ Hence, the given condition reduces to
$\begin{align*} |a-h|^2=2|a|^2&=2 \\ (a-h)\overline{(a-h)}&=2 \\ \frac{1}{2}\left(a-b-c+\frac{bc}{a}\right)\overline{\frac{1}{2}\left(a-b-c+\frac{bc}{a}\right)}&=2 \\ \left(a-b-c+\frac{bc}{a}\right)\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}+\frac{a}{bc}\right)&=8 \\ \frac{a^2}{bc}+\frac{bc}{a^2}-\frac{2a}{b}-\frac{2b}{a}-\frac{2a}{c}-\frac{2c}{a}+\frac{b}{c}+\frac{c}{b}&=4. \end{align*}$ Then, by complex foot again, we have
$p=\frac{1}{2}\left(a+b+h-ab\overline{h}\right),$ which expands to
$\begin{align*} p&=\frac{1}{4}\left(2a+2b+\left(a+b+c-\frac{bc}{a}\right)-ab\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}\right)\right) \\ &=\frac{1}{4}\left(2a+2b+c-\frac{bc}{a}-\frac{ab}{c}+\frac{a^2}{c}\right). \end{align*}$ Analogously, we have
$q=\frac{1}{4}\left(2a+2c+b-\frac{bc}{a}-\frac{ac}{b}+\frac{a^2}{b}\right).$ By the complex collinearity criterion, it suffices to show that
$\frac{0-p}{0-q}=\frac{p}{q} \in \mathbb{R}.$ Thus, we calculate
$\begin{align*} \frac{p}{q}&=\frac{2a+2b+c-\frac{bc}{a}-\frac{ab}{c}+\frac{a^2}{c}}{2a+2c+b-\frac{bc}{a}-\frac{ac}{b}+\frac{a^2}{b}} \\ &=\frac{2+\frac{2b}{a}+\frac{c}{a}-\frac{bc}{a^2}-\frac{b}{c}+\frac{a}{c}}{2+\frac{2c}{a}+\frac{b}{a}-\frac{bc}{a^2}-\frac{c}{b}+\frac{a}{b}}. \end{align*}$ Now, let $x$ and $y$ be the numerator and denominator of this fraction, respectively, so that $\frac{p}{q}=\frac{x}{y}$ . Note that
$x+\overline{x}=4+\left(-\frac{a^2}{bc}-\frac{bc}{a^2}+\frac{2a}{b}+\frac{2b}{a}+\frac{2a}{c}+\frac{2c}{a}-\frac{b}{c}-\frac{c}{b}\right)=4-4=0$ by substituting directly into the result we derived from the given condition. Similarly, we get $y+\overline{y}=0$ . Hence,
$\frac{x}{y}=\frac{-\overline{x}}{-\overline{y}}=\frac{\overline{x}}{\overline{y}} \implies \frac{x}{y} \in \mathbb{R},$ which clearly finishes.

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blueprimes

362 posts

blueprimes

#115 h Aug 13, 2024, 2:18 PM

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Note that $APHQ$ is cyclic since $\angle APH + \angle HQA = 90^\circ + 90^\circ = 180^\circ$ , and inscribed angles and AA Similarity easily shows $\triangle APQ \sim ACB$ (inversely similar.) Now
$\angle AOP = 180^\circ - \angle PAO - \angle APO = 180^\circ - \angle CAH - \angle AHQ = 180^\circ - 90^\circ = 90^\circ.$ Therefore, $AO \perp PQ$ , so it is the altitude of $\triangle APQ$ with respect to $A$ . Then
$\dfrac{AO}{AH} = \dfrac{AQ}{AB} \implies \dfrac{c}{2b \sin(C)^2} = \dfrac{b \sin(C)^2}{c} \implies \dfrac{c}{2b \sin(C)^2} = \dfrac{1}{\sqrt{2}}$ which yields $AH = AO \sqrt{2} \implies AH^2 = 2 AO^2$ as desired.

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maths_enthusiast_0001

133 posts

maths_enthusiast_0001

#116 h Oct 12, 2024, 11:30 PM

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Easy for a JMO P5 Geometry.
Toss the figure on the coordinate plane. Set $A \equiv (1,b), B \equiv(0,0), C \equiv (a,0)$ and $H \equiv (1,0)$ . Now to find coordinates of $P,Q$ we can use:
Lemma: Let the foot of perpendicular from a point $(x_{0},y_{0})$ to a line $ax+by+c=0$ be $(h,k)$ . Then,
$\boxed{\dfrac{h-x_{0}}{a}=\dfrac{k-y_{0}}{b}=-\left(\dfrac{ax_{0}+by_{0}+c}{a^{2}+b^{2}}\right)}.$ Thus we shall get, $\boxed{P \equiv \left(\dfrac{1}{b^{2}+1},\dfrac{b}{b^{2}+1}\right)}$ and $\boxed{Q \equiv \left(\dfrac{a^{2}-2a+ab^{2}+1}{a^{2}-2a+b^{2}+1},\dfrac{a^{2}b-2ab+b}{a^{2}-2a+b^{2}+1}\right)}$ .
Now we can find the coordinates of $O$ by intersection of perpendicular bisectors of sides. Thus,
$\boxed{O \equiv \left(\dfrac{a}{2},\dfrac{b^{2}-a+1}{2b}\right)}$ Now clearly we have $AH^{2}=b^{2}$ and $AO^{2}=\left(\dfrac{a}{2}-1\right)^{2}+\left(\dfrac{b^{2}-a+1}{2b}-b\right)^{2}$ . Thus,
$AH^{2}=2AO^{2}$ $\implies b^{2}=2\left(\left(\dfrac{a}{2}-1\right)^{2}+\left(\dfrac{b^{2}-a+1}{2b}-b\right)^{2}\right)$ $\implies 2b^{4}=b^{4}+2b^{2}+1+a^{2}-2a+a^{2}b^{2}-2ab^{2}$ $\implies b^{4}-2b^{2}-1-a^{2}+2a-a^{2}b^{2}+2ab^{2}=0$ $\implies \boxed{b^{2}(b^{2}-a^{2}+2a-2)=(a-1)^{2}}$ Now we need to find the equation of line $PQ$ and satisfy the coordinates of $O$ in it to get the condition for $O,P$ and $Q$ to be collinear.
Let $m_{PQ}$ be the slope of line $PQ$ . Then,
$m_{PQ}=\left(\dfrac{\dfrac{b}{b^{2}+1}-\dfrac{a^{2}b-2ab+b}{a^{2}-2a+b^{2}+1}}{\dfrac{1}{b^{2}+1}-\dfrac{a^{2}-2a+ab^{2}+1}{a^{2}-2a+b^{2}+1}}\right)$ $\implies m_{PQ}=\dfrac{b^{3}(-2a+a^{2})}{a^{2}b^{2}-ab^{2}+ab^{4}}$ $\implies m_{PQ}=\dfrac{b(a-2)}{a+b^{2}-1}$ Now we can find the equation of line $PQ$ as:
$\boxed{y=\left(\dfrac{b(a-2)}{a+b^{2}-1}\right)x+\left(\dfrac{b}{a+b^{2}-1}\right)}$ Now we can just plug the coordinates of $O$ into it to get the desired condition. Thus for $O,P,Q$ to be collinear we must have,
$\dfrac{b^{2}-a+1}{2b}=\dfrac{\dfrac{ab(a-2)}{2}+b}{a+b^{2}-1}$ $\implies 2b^{2}+b^{2}(a^{2}-2a)=b^{4}-(a-1)^{2}$ $\implies \boxed{b^{2}(b^{2}-a^{2}+2a-2)=(a-1)^{2}}.$ This condition is exactly equivalent to that of $AH^{2}=2AO^{2}$ and hence, we are done. $\blacksquare$ :coolspeak:

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eg4334

636 posts

eg4334

#117 h Oct 13, 2024, 4:05 AM

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The condition we want to prove is equivalent to $\angle AQO = \angle PQA$ or $\angle AQO = \angle PHA$ by a cyclic quadrilateral. Because $AH$ and $AO$ are isogonal in $A$ , this reduces to proving that $\triangle APH \sim \triangle AOQ$ or eqiuvalently $\triangle AOQ \sim AHB$ ergo $\frac{AH^2}{AO^2} = \frac{AB^2}{AQ^2}$ $2 = \frac{AB^2}{AQ^2}$ So from our condition we just need to prove $2 AQ^2 = AB^2$ and we are done.

Now we resort to trig. We have that $AO \sin{\angle C} = \frac{AB}{2}$ and $AH \sin{\angle C} = AQ$ by basic right triangles. Plugging this into our given condition we get $2 \cdot \left( \frac{AB}{2 \sin{\angle C}} \right)^2 = \left( \frac{AQ}{\sin{\angle C}} \right)^2$ which reduces to $AB^2 = 2AQ^2$ , done.

This post has been edited 1 time. Last edited by eg4334, Dec 28, 2024, 12:07 AM

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bjump

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bjump

#118 h Feb 26, 2025, 6:10 PM • 1 Y

Y by imagien_bad

Bh this is fake geo.

Let $A'$ be the antipode of $A$ . Then we have $APHQ \sim ACA'B$ . Note that $AH = \sqrt{2}R = c \sin(b) = b \sin(c) = 2R\sin(b)\sin(c) \implies \sin(b)\sin(c) = \frac{\sqrt{2}}{2}.$ Note that $\frac{AQ}{AO} =\frac{\sqrt{2}R \sin(c)}{R} = \frac{1}{\sin(b)}$ Therefore by right triangle ratios $\angle AOQ = 90^\circ$ , symmetry gives $\angle AOP = 90^\circ$ therefore $\angle POQ= \angle POA + \angle AOQ= 180^\circ$ . Done.

This post has been edited 1 time. Last edited by bjump, Feb 26, 2025, 6:16 PM
Reason: typo

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