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Source: Indonesian National Mathematical Olympiad 2024, Problem 3

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somebodyyouusedtoknow

259 posts

somebodyyouusedtoknow

#1 h Aug 28, 2024, 7:03 AM • 2 Y

Y by LuoJi, Rounak_iitr

The triangle $ABC$ has $O$ as its circumcenter, and $H$ as its orthocenter. The line $AH$ and $BH$ intersect the circumcircle of $ABC$ for the second time at points $D$ and $E$ , respectively. Let $A'$ and $B'$ be the circumcenters of triangle $AHE$ and $BHD$ respectively. If $A', B', O, H$ are not collinear, prove that $OH$ intersects the midpoint of segment $A'B'$ .

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#2 h Aug 28, 2024, 7:41 AM • 1 Y

Y by LuoJi

We claim $A'B'OH$ is a parallelogram. Spiral sim at $A$ sending $A'O$ to $BH$ tells us that $A'O=BH\times\frac{AO}{AB}$ and spiral sim at $B$ sending $B'H$ to $AO$ gives $B'H=AO\times\frac{BH}{AB}$ clearly the two expressions are equal hence $A'O=B'H$ similarly we get opposite pair of equal sides yay.

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#3 h Aug 28, 2024, 8:03 AM • 1 Y

Y by LuoJi

Construction of $O,H,D,E$ motivate to use complex :maybe:

:maybe:

W.L.O.G. $O=0$ and $(ABC)$ as a unit circle. We have $h=a+b+c$ dan $|a|=|b|=|c|=|d|=|e|=1$ (imply $\overline{a}=1/a$ by $a\overline{a}=|a|^2$ property). Because $BE\;\bot\;AC$ we have $be+ac=0\iff e = -\frac{ac}{b}$ and $d=-\frac{bc}{a}$ . Solve the system equations for $a'$ :
$|a'-a| = |a'-h| = |a'-e|,$ we have
$\begin{align*} a' &= \frac{\left |\begin{matrix} h &h\overline{h} & 1\\ a & a\overline{a} & 1 \\ e&e\overline{e} & 1 \end{matrix} \right |}{\left |\begin{matrix} h &\overline{h} & 1\\ a & \overline{a} & 1 \\ e&\overline{e} & 1 \end{matrix} \right |} = \frac{\left |\begin{matrix} a+b+c &(a+b+c)\left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right )&1 \\ a & 1 & 1\\ -\frac{ac}{b} & 1 & 1 \end{matrix} \right |}{\left |\begin{matrix} a+b+c & \frac{1}{a} + \frac{1}{b} + \frac{1}{c} & 1\\ a & \frac{1}{a} & 1 \\ -\frac{ac}{b} & -\frac{b}{ac} & 1 \end{matrix} \right |}\\ &= \frac{-\frac{(a+b)(a+c)(b+c)^2}{b^2c}}{\frac{(b-a)(b+a)(b+c)^2}{ab^2c}} = -\frac{(a+c)a}{b-a}. \end{align*}$ Analogously, $b' = -\frac{(b+c)b}{a-b}.$ Therefore, the midpoint of $A'B'$ is
$m = \frac{a'+b'}{2} = \frac{1}{2}\left (\frac{-a^2-ac +b^2+bc}{b-a} \right )=\frac{(b-a)(a+b+c)}{2(b-a)}=\frac{a+b+c}{2},$ which is also the midpoint of $OH$ .

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#4 h Aug 28, 2024, 8:11 AM • 2 Y

Y by godjuansan, LuoJi

Here is an alternate but similar way to prove the parallelogram claim: Note that $AE$ is the radical axis of $\odot(AHE)$ and $\odot(ABC)$ , so $OA' \perp AE$ . Also:
$\begin{align*} \measuredangle(B'H, AE) &= \measuredangle(B'H, HB) + \measuredangle(BE, EA) \\ &\stackrel{\triangle HB'B \sim \triangle AOB}{=} \measuredangle(OA, AB) + \measuredangle(BC, CA) \\ &\stackrel{AO \text{ and } AD \text{ are isogonal w.r.t. } \angle BAC}{=} \measuredangle(CA, AD) + \measuredangle(BC, CA) = \measuredangle(BC, AD) = 90^{\circ}. \end{align*}$ So, $B'H \perp AE$ as well. Therefore, $B'H \parallel OA'$ . Similarly, $A'H \parallel OB'$ , proving the claim.

This post has been edited 1 time. Last edited by Seicchi28, Aug 28, 2024, 8:13 AM

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crocodilepradita

145 posts

crocodilepradita

#5 h Aug 28, 2024, 8:49 AM • 2 Y

Y by TheoSi, LuoJi

From PoP on B' we can chase the length of B'O using trigonometry, same with A'H etc. Hence we can trigon bash this problem

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#6 h Aug 28, 2024, 11:39 AM • 2 Y

Y by Seicchi28, LuoJi

First notice that triangle $AHE$ and $BHD$ are isosceles triangle and thus $A'$ and $B'$ lies on $AC$ and $BC$ respectively. Let $P, Q$ be midpoint of $AH$ and $BH$ respectively. Let $S_a = \frac{-a^2 + b^2 + c^2}{2}$ , define $S_b$ and $S_c$ similarly. We are going to use Barycentric Coordinates.

Let ABC be reference triangle and $A(1, 0, 0); B(0, 1, 0); C(0, 0, 1)$ . Well known that $H(S_bS_c : S_cS_a : S_aS_b)$ and $O(a^2S_a : b^2S_b : c^2 S_c)$ . Let $T = S_bS_c + S_cS_a + S_aS_b = \frac{1}{2} (a^2S_a + b^2S_b + c^2 S_c)$ .

We can compute that $Q(S_bS_c : S_cS_a + T : S_aS_b)$ and similarly $P(S_bS_c + T : S_c S_a : S_aS_b)$ .

Next, we can compute $A'$ by using the fact that $PA' // BC$ . Hence, $P$ , $A$ , and $R$ are collinear where $R(0 : 1 : -1)$ is point of infinity of line $BC$ .

$\begin{vmatrix} x & 0 & z \\ 0 & 1 & -1\\ S_bS_c + T & S_cS_a & S_aS_b \end{vmatrix} = 0$
Expanding, we can get $A'(c^2S_c + b^2S_b : 0 : a^2S_a)$ . Similarly, $B'(0 : a^2S_a + c^2S_c : b^2S_b)$ .
Let $N$ be the midpoint of $A'B'$ , we can see that $N(c^2S_c + b^2S_b : a^2S_a + c^2S_c : a^2S_a + b^2S_b)$ .

Now, we just need to prove that $O, N, H$ are collinear (and in fact $N$ is the midpoint of $OH$ ), that is:

$\begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ b^2S_b + c^2S_c & c^2S_c + a^2S_a & a^2S_a + b^2S_b\\ a^2S_a & b^2S_b & c^2S_c \end{vmatrix} = \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ a^2S_a + b^2S_b + c^2S_c & c^2S_c + a^2S_a + b^2S_b & a^2S_a + b^2S_b + c^2S_c\\ a^2S_a & b^2S_b & c^2S_c \end{vmatrix} = 2T \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ 1 & 1 & 1\\ S_aS_b + S_aS_c & S_aS_b + S_bS_c & S_aS_c + S_bS_c \end{vmatrix} = 2T \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ 1 & 1 & 1\\ T & T & T \end{vmatrix} = 0$ which is evident from the last determinant.

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#7 h Aug 28, 2024, 11:50 AM • 1 Y

Y by LuoJi

Notice that the segment $AE$ is the reflection of $AH$ with respect to the line $AC$ . Thus, if we let $X$ be the point on $AC$ such that $AH \bot HX$ , the point $A'$ is the midpoint of $AX$ . The construction of $X$ also means $HX$ is parallel to $BC$ . Similarly, $B'$ is the midpoint of $BY$ , where $Y$ is the point on $BC$ such that $HY$ is parallel to $AC$ . Thus, the midpoint of $A'B'$ is just the centroid of $4$ points: $X$ , $Y$ , $A$ , and $B$ .

Now notice that $XCYH$ is a parallelogram, so the midpoint of $XY$ is equal to the midpoint of $CH$ . Then the midpoint of $A'B'$ is the centroid of $4$ points $A$ , $B$ , $C$ , and $H$ . Using complex coordinate or some other way, it is now easy to see that this centroid is actually the midpoint of $HO$ .

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#8 h Aug 28, 2024, 2:47 PM • 3 Y

Y by BlazingMuddy, Nartku, LuoJi

godjuansan wrote:

First notice that triangle $AHE$ and $BHD$ are isosceles triangle and thus $A'$ and $B'$ lies on $AC$ and $BC$ respectively. Let $P, Q$ be midpoint of $AH$ and $BH$ respectively. Let $S_a = \frac{-a^2 + b^2 + c^2}{2}$ , define $S_b$ and $S_c$ similarly. We are going to use Barycentric Coordinates.

Let ABC be reference triangle and $A(1, 0, 0); B(0, 1, 0); C(0, 0, 1)$ . Well known that $H(S_bS_c : S_cS_a : S_aS_b)$ and $O(a^2S_a : b^2S_b : c^2 S_c)$ . Let $T = S_bS_c + S_cS_a + S_aS_b = \frac{1}{2} (a^2S_a + b^2S_b + c^2 S_c)$ .

We can compute that $Q(S_bS_c : S_cS_a + T : S_aS_b)$ and similarly $P(S_bS_c + T : S_c S_a : S_aS_b)$ .

Next, we can compute $A'$ by using the fact that $PA' // BC$ . Hence, $P$ , $A$ , and $R$ are collinear where $R(0 : 1 : -1)$ is point of infinity of line $BC$ .

$\begin{vmatrix} x & 0 & z \\ 0 & 1 & -1\\ S_bS_c + T & S_cS_a & S_aS_b \end{vmatrix} = 0$
Expanding, we can get $A'(c^2S_c + b^2S_b : 0 : a^2S_a)$ . Similarly, $B'(0 : a^2S_a + c^2S_c : b^2S_b)$ .
Let $N$ be the midpoint of $A'B'$ , we can see that $N(c^2S_c + b^2S_b : a^2S_a + c^2S_c : a^2S_a + b^2S_b)$ .

Now, we just need to prove that $O, N, H$ are collinear (and in fact $N$ is the midpoint of $OH$ ), that is:

$\begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ b^2S_b + c^2S_c & c^2S_c + a^2S_a & a^2S_a + b^2S_b\\ a^2S_a & b^2S_b & c^2S_c \end{vmatrix} = \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ a^2S_a + b^2S_b + c^2S_c & c^2S_c + a^2S_a + b^2S_b & a^2S_a + b^2S_b + c^2S_c\\ a^2S_a & b^2S_b & c^2S_c \end{vmatrix} = 2T \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ 1 & 1 & 1\\ S_aS_b + S_aS_c & S_aS_b + S_bS_c & S_aS_c + S_bS_c \end{vmatrix} = 2T \begin{vmatrix} S_bS_c & S_cS_a & S_aS_b \\ 1 & 1 & 1\\ T & T & T \end{vmatrix} = 0$ which is evident from the last determinant.

name a more iconic duo than godjuansan and barycentric coordinates, i'll wait. (bro's only on aops now for posting bary solutions :coolspeak:

:coolspeak:

)

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Nartku

#9 h Aug 31, 2024, 5:07 PM • 1 Y

Y by LuoJi

By reflecting the orthocenter, we can clearly see that A' lies on AC and B' lies on BC. Notice that A' is the intersection of the perpendicular bisector of AH with line AC and B' is the intersection of the perpendicular bisector of BH with line BC, therefore we can simply use cartesian coordinates to prove that the midpoint of OH and midpoint of A'B' coincide

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268 posts

iStud

#10 h Sep 2, 2024, 7:57 AM • 1 Y

Y by LuoJi

To generalize the problem, we use directed angle mod $180$ .
The main idea here is to prove that $A'OB'H$ is a parallelogram. To prove that, we can simply prove that $\triangle{A'OE}$ is congruent to $\triangle{B'OD}$ . To do this, try to prove 2 of the 3 angles are the same and use the fact that $OD=OE$ , then the conclusion follows from that.

This post has been edited 1 time. Last edited by iStud, Sep 2, 2024, 7:58 AM

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#12 h Sep 2, 2024, 8:50 AM • 1 Y

Y by LuoJi

Proof without word. (with the knowledge that $\triangle AHE$ and $\triangle{BDH}$ are isosceles)

Attachments:

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#13 h Sep 2, 2024, 9:19 AM

Y by

KHOMNYO2 wrote:

Proof without word. (with the knowledge that $\triangle AHE$ and $\triangle{BDH}$ are isosceles)

The diagram also holds for obtuse, for anyone wondering how does my food taste (even though it looked quite different). Even though, one diagram is enough but i had the urge to add its obtuse version too, lol

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Reason: .

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268 posts

iStud

#14 h Sep 4, 2024, 10:00 PM

Y by

Btw, here is my full solution. I solved this problem by doing the case when $\triangle{ABC}$ is acute, the obtuse one can be proved similarly.

It is well-known that $\triangle{AHE}$ and $\triangle{BHD}$ are isosceles triangles and also $O$ and $H$ are isogonal conjugates w.r.t the angles of $\triangle{ABC}$ . Because of this, clearly $A'$ and $B'$ lies on $AC$ and $BC$ . Note that $\angle{CA'E}=2\cdot\angle{A'AE}=2\cdot\angle{CAE}=\angle{COE}$ and similarly $\angle{CB'D}=2\cdot\angle{B'BD}=2\cdot\angle{CBD}=\angle{COD}$ $\Longrightarrow$ $A'OCE$ and $B'OCD$ are cyclic quads.

Now by considering the fact that $OD=OE$ and $\angle{A'EO}=\angle{A'CO}=\angle{ACO}=\angle{CAO}=\angle{BAH}=\angle{BAD}=\angle{BCD}=\angle{B'CD}=\angle{B'OD}$ and $\angle{A'OE}=\angle{A'CE}=\angle{ACE}=\angle{ABE}=\angle{ABH}=\angle{OBC}=\angle{OCB}=\angle{OCB'}=\angle{B'DO}$ , thus $\triangle{A'OE}$ is congruent to $\triangle{B'OD}$ . For the final act, we have $B'H=B'D=A'O$ and $A'H=A'E=B'O$ , hence $A'OB'H$ is a parallelogram and so $OH$ bisects $A'B'$ . $\blacksquare$

Actually, I have another solution that doesn't requires any additional cyclic quads, but it's written in Indonesian so I hope you guys can read it

Attachments:

P3_OSN_2024___Day_1_compressed.pdf (66kb)

This post has been edited 5 times. Last edited by iStud, Sep 8, 2024, 7:17 AM

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#15 h Sep 11, 2024, 7:44 AM

Y by

This is my problem. One of two official solutions has been posted, using complex. Another solution similar to @above (use directed angle)

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#16 h Apr 29, 2025, 2:48 AM

Y by

Wrote this sol in the middle of an informatics class. Edit: I just realized that I switched up $A'$ and $B'$ lol.
We shall use trig bash
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.133476424526859, xmax = 15.371992371922431, ymin = -1.1681467670540358, ymax = 11.582353551851757; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw(circle((5,5), 5), linewidth(0.4) + zzttqq); draw((0.7001449911989663,2.4482854973001658)--(9.299855008801034,2.4482854973001658), linewidth(0.4)); draw((3.4419996538801927,9.751066714064386)--(0.7001449911989663,2.4482854973001658), linewidth(0.4)); draw((3.4419996538801927,9.751066714064386)--(9.299855008801034,2.4482854973001658), linewidth(0.4)); draw((0.7001449911989663,2.4482854973001658)--(3.4419996538801914,4.647637708664716), linewidth(0.4)); draw((3.4419996538801914,4.647637708664716)--(3.4419996538801914,0.2489332859356157), linewidth(0.4)); draw((3.4419996538801914,0.2489332859356157)--(0.7001449911989663,2.4482854973001658), linewidth(0.4)); draw((3.4419996538801927,9.751066714064386)--(3.4419996538801914,4.647637708664716), linewidth(0.4)); draw((3.4419996538801914,4.647637708664716)--(8.423858873816904,8.643787920857257), linewidth(0.4)); draw((8.423858873816904,8.643787920857257)--(3.4419996538801927,9.751066714064386), linewidth(0.4)); draw(circle((2.9531668798243818,2.4482854973001658), 2.2530218886254154), linewidth(0.4) + fuqqzz); draw(circle((5.48883277405581,7.199352211364552), 3.271203559110397), linewidth(0.4) + qqwuqq); draw((2.9531668798243818,2.4482854973001658)--(5.48883277405581,7.199352211364552), linewidth(0.4)); draw((5.48883277405581,7.199352211364552)--(5,5), linewidth(0.4)); draw((2.9531668798243818,2.4482854973001658)--(3.4419996538801914,4.647637708664716), linewidth(0.4)); draw((2.9531668798243818,2.4482854973001658)--(5,5), linewidth(0.4)); draw((3.4419996538801914,4.647637708664716)--(5.48883277405581,7.199352211364552), linewidth(0.4)); draw((3.4419996538801914,4.647637708664716)--(5,5), linewidth(0.4)); draw((5,5)--(5,2.4482854973001658), linewidth(0.4)); draw((5,5)--(6.370927331340614,6.099676105682276), linewidth(0.4)); /* dots and labels */ dot((5,5),linewidth(3pt) + dotstyle); label("$O$", (5.063752064349315,5.0960915737019175), NE * labelscalefactor); dot((3.4419996538801927,9.751066714064386),linewidth(3pt) + dotstyle); label("$A$", (3.509586602592138,9.85374094642796), NE * labelscalefactor); dot((0.7001449911989663,2.4482854973001658),linewidth(3pt) + dotstyle); label("$B$", (0.7660087976534472,2.542819743672275), NE * labelscalefactor); dot((9.299855008801034,2.4482854973001658),linewidth(3pt) + dotstyle); label("$C$", (9.361495331045184,2.542819743672275), NE * labelscalefactor); dot((3.4419996538801914,4.647637708664716),linewidth(3pt) + dotstyle); label("$H$", (3.509586602592138,4.747197286368674), NE * labelscalefactor); dot((3.4419996538801914,0.2489332859356157),linewidth(3pt) + dotstyle); label("$D$", (3.509586602592138,0.33844220097587574), NE * labelscalefactor); dot((8.423858873816904,8.643787920857257),linewidth(3pt) + dotstyle); label("$E$", (8.489259612712074,8.74362275945855), NE * labelscalefactor); dot((2.9531668798243818,2.4482854973001658),linewidth(3pt) + dotstyle); label("$A'$", (3.0179628340771125,2.542819743672275), NE * labelscalefactor); dot((5.48883277405581,7.199352211364552),linewidth(3pt) + dotstyle); label("$B'$", (5.555375832864341,7.300469116398317), NE * labelscalefactor); dot((4.220999826940097,4.82381885433236),linewidth(3pt) + dotstyle); label("$N_{9}$", (4.286669333470726,4.921644430035296), NE * labelscalefactor); dot((5,2.4482854973001658),linewidth(3pt) + dotstyle); label("$M$", (5.063752064349315,2.542819743672275), NE * labelscalefactor); dot((6.370927331340614,6.099676105682276),linewidth(3pt) + dotstyle); label("$N$", (6.427611551197451,6.190350929428908), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
The big claim is that $B'OA'H$ is a parallelogram, since the diagonals of a parallelogram intersect
at the midpoints, we are done. To do this it suffices to show that $A'H = OB'$ , by symmetry
it follows that $A'O = HB'$ , thus we would be done. Define $N$ as the midpoint of $AC$ and $M$
as the midpoint of $BC$ .

Notice that by the Pythagorean theorem we know that
$OB'^2 = ON^2 + {(AN - AB')}^2 = ON^2 + AN^2 - 2AN \cdot AB + AB'^2 = R^2 - 2AN \cdot AB' + AB'^2$ We shall calculate the value of $BA'$ , notice that since $\angle BHC = 180 - \angle BAC$ , it follows that
the circumradius of $(BHC)$ is equal to that of $(ABC)$ , thus by law of sines on $\triangle BHC$
$\frac{BH}{\sin(90 - \beta)} = 2R \Longrightarrow 2R\cos(\beta) = BH$ We know that the circumradius of $(BHD)$ is equal to $A'H = A'B$ , thus we know that
$A'B = \frac{BH}{2\sin(\angle BDH)} = \frac{BH}{2\sin(\gamma)} = \frac{R\cos(\beta)}{\sin(\gamma)}$ Similarly we find that
$AB' = \frac{R\cos(\alpha)}{\sin(\gamma)}$ We wish to show that $OB' = A'H$ , however $A'H = A'B$ , thus substituting variables we wish to show
$A'B^2 = OB'^2 \iff \frac{R^2\cos^2(\beta)}{\sin^2(\gamma)} = R^2 - 2AN \cdot AB' + AB'^2$ We know that
$AN = \frac{AC}{2} = R\sin(\beta)$ Thus we need to show
$\frac{R^2\cos^2(\beta)}{\sin^2(\gamma)} = R^2 - 2(R\sin(\beta)) \cdot \left(\frac{R\cos(\alpha)}{\sin(\gamma)}\right) + {\left(\frac{R\cos(\alpha)}{\sin(\gamma)}\right)}^2$ which is equivalent to showing
$\frac{\cos^2(\beta)}{\sin^2(\gamma)} = 1 - \frac{2\sin(\beta)\cos(\alpha)}{\sin(\gamma)} + \frac{\cos^2(\alpha)}{\sin^2(\gamma)} \iff \cos^2(\beta) = \sin^2(\gamma) - 2\sin(\beta)\sin(\gamma)\cos(\alpha) + \cos^2(\alpha)$ We can sub $\sin(\gamma) = \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)$ to show that it suffices to prove
$\cos^2(\beta) - \cos^2(\alpha) = [\sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)][\sin(\alpha)\cos(\beta) - \sin(\beta)\cos(\alpha)]$ But this is true since it is equivalent to
$\cos^2(\beta) - \cos^2(\alpha) = \sin^2(\alpha)\cos^2(\beta) - \sin^2(\beta)\cos^2(\alpha)$ Which is equivalent to
$\cos^2(a)\cos^2(\beta) = \cos^2(\beta)\cos^2(\alpha)$

This post has been edited 1 time. Last edited by Ihatecombin, Apr 29, 2025, 6:41 AM
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