simple angle chase

Can we please make this the official thread, I spent 30 minutes thinking of the title Anyway, two steps: 1. Prove $FB=FE.$ 2. Prove $ADEOF$ cyclic where $O$ is the circumcenter. Both super straightforward.

No config issues right?

cmsgr8er wrote: No config issues right? nope

who is dennis

franchester wrote: who is dennis hi

lol trivial by angle chasing

Triangle based problem statement in case anybody wants it: Let $ABC$ be a triangle, and let $D$ be the midpoint of minor arc $BC$. $E$ and $F$ are chosen on $AC$ and $DB$ such that $DE \perp BC$ and $EF || DC$. Prove that $FA=FD$. Also,

I thought this problem was a bit silly. Since $\overline{EF} \parallel \overline{BC}$, $ADEF$ is cyclic. Then $\angle DEA = \angle DFA$ and $\angle DCA = \angle DBA$ so \[ \triangle AEC \sim \triangle DFB. \]Clearly $EA = EC$, so $FD = FB$.

which problem was this, on which test?

This is USOJMO #4 50th post!

jmo 2020/4 (edited source in post)

Can someone check my sol? Do you think it will receive any points?

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Is this the official thread? When you answer "Trivial by geometry." on USAJMO... We wish to prove that $\triangle FBD$ is isosceles. It suffices to prove that that $\angle BFD=\angle DBF.$ Lemma. Points $A,D,E,F$ form a cyclic quadrilateral. Proof. Opposite angles add to $180^{\circ}. (\angle ABC=\angle EFA, \text{which implies that the opposite angles add to 180 degrees}). ~ \blacksquare$ Connect the diagonals of quadrilateral $ADEF.$ We get $\angle AFD=\angle AED, \angle ACD=\angle ABD \implies \angle BFD=\angle DBF,$ and the desired result follows. $~ \blacksquare$ ... Then you realize it really is trivial by geometry. (oR cOmPlEx BaSh) Lol, is this the solution everybody had, nobody complex bashed?

Doesnt seem like it...

First off, note that $BE$ is the perpendicular bisector of $\overline{AC},$ as $AB = AC.$ This means that $EA = EC.$ Now, we claim that $\triangle DFB$ and $\triangle AEC$ are oppositely similar. We will show this using angle-chasing. For the first angle of our similarity, we know that $$\measuredangle FBD = \measuredangle ABD = \measuredangle ACD = \measuredangle ACE = -\measuredangle ECA,$$from the given cyclicity of $ABCD.$ Now, it is not too hard to see that $$\measuredangle DAF = \measuredangle DAB = \measuredangle DCB = \measuredangle DEF$$since $EF \parallel BC,$ implying that $ADEF$ is cyclic. This then means that $$\measuredangle DFB = \measuredangle DFA = \measuredangle DEA = \measuredangle CEA = -\measuredangle AEC.$$Therefore, $\triangle DFB$ and $\triangle AEC$ are oppositely similar, as claimed. As a consequence, since $AE = EC,$ we have that $FB = FD,$ as desired.

We can see that $AE = EC$ This is because $EB$ includes the center of the circle. We also know that $AFED$ is cyclic because $FE \parallel BC$. Since $ABCD$ is cyclic we know that $\angle BAC = \angle BDC$. Combining we see that $\angle FBD = \angle ACD = \angle CAE = \angle FDB$. This give the result. $\blacksquare$

Pretty straightforward, solved in like 15 minutes. Here's a solution without some details. Define $\angle{DBF} = \alpha$ and $\angle{CAB} = \beta$. Note that \[\angle{DAF} =180-\angle{ECB} = 180-\angle{DEF}\]which implies that $DEFA$ is cyclic. Also note that from circle properties, $\angle{DAF} = 180-(\alpha+\beta)$. We also have from the fact that $EB$ is the perpendicular bisector of $AC$, that $\angle{EAF} = \alpha+\beta$. Thus \[\angle{EFD} = \angle{EAD} = \angle{DAE}-\angle{EAF} = 180-2(\alpha+\beta)\]From parallel lines, we have \[\angle{EFD} = \alpha+\beta \implies \angle{EDF} = \alpha+\beta \implies \angle{BDF} = \alpha = \angle{DBF} \implies DF = BF\]

bassicaly we hav ADEF is cyclic so aec is similar to dfb so ea=ce so df=bf or you can soivle withe simpel angle chasing yay my fifth day 2 jmo solve

Let $\angle BCA=a, \angle ACD = b$ Let the intersection point of $BE$ and $AC$ be $Z$ Let the intersection point of $BD$ and $AC$ be $Y$ By same arc, $\angle BDA = a, \angle ABD = b$ So $\angle BAD = 180-a-b$ By parallel lines, $\angle CYE=a$, so $\angle FED = a+b$, so $AFED$ is cyclic $\triangle BAZ$ and $\triangle BCZ$ are congruent by HL so $AZ=CZ$ and $\triangle AZE$ and $\triangle CZE$ are congruent by SAS So $AE=CE$ So $\angle CAE = b$ and $\angle AED = 2b$ So $\angle FEA = a-b$ By $AFED$ cyclic, $\angle FDA = a-b$ So $\angle FDB = b$ And because $\angle FBD = b$, $FB=FD$

Let $\alpha=\angle BAC$ and $\beta=\angle ACD$. Then \[\angle ADC=2\alpha=\angle BFE,\]so $F,A,E,D$ are cyclic. Then \[\alpha+\beta=\angle BCE=\angle BAE=\angle EDF=\angle BDF+\angle BDC=\angle BDF+\alpha,\]so \[\angle BDF=\beta=\angle ACD=\angle ABD=\angle FBD\;\blacksquare\]

By Reim's theorem, note that since $EF\parallel BC$, we have that since $ABCD$ is cyclic, so is $AFED$. Now note that since $ABCD$ is cyclic, we have that \[\angle FBD=\angle ACD.\]Additionally, since $AB=BC$ and $BE\perp AC$, this means that $AE=EC$, which means that \[\angle AED=\angle ACE+\angle CAE=2\angle ACD,\]and since $AFED$ is cyclic, this gives that \[\angle FDB=\angle AFD-\angle FBD=\angle AED-\angle ACD,\]which is equal to $\angle ACD$, since $\angle AED=2\angle ACD$. Using this, we get that $\angle FBD=\angle FDB=\angle ACD$, meaning that $\triangle FBD$ is isosceles with $FB=FD$, finishing the problem. *Note: something cool I also found is that we can actually get $FE=FB=FD$?

Let $(ABC)=\Gamma , \angle ACB=\alpha , \angle ABD=\beta , BE \cap AC=\{O\} $ $\angle ABD \stackrel{\Gamma }{=} \angle ACD=\beta$ From triangle $\triangle EOC$ we get : $\angle CEO=90-\alpha$ From triangle $\triangle BOC$ we get : $\angle CBO \equiv \angle CEO =90-\beta$ $EF \parallel BC \implies \angle FEB=\angle CBE=90-\alpha \implies \angle FEB=90-\alpha$ $\textbf{Claim:}$ $\angle AEO=\angle CEO$

So: $\angle AEO=\angle CEO=90-\beta \implies \angle AEO=90-\beta$ $\angle AEO=\angle AEF+ \angle FEB \implies \angle AEF=\alpha -\beta$ $\textbf{Claim:}$ The quadrilateral $\square ADEF-is$ $cyclic$

Let $(ADEF)=\omega$ $\angle ADF \stackrel{\omega}{=} \angle AEF=\alpha-\beta \implies \angle ADF=\alpha-\beta$ $\alpha=\angle ACB \stackrel{\Gamma }{=}\angle ADB=\angle ADF + \angle FDB \implies \angle FDB=\beta$ $\angle FDB = \beta= \angle ABD \equiv \angle FBD \implies \angle FDB =\angle FBD \implies FB=FD$ As desired $\blacksquare$

Angle chasing, we have that \[ \angle FBD = \angle ABD = \angle ACD = \angle ACE. \]In addition, \[ \angle FED + \angle FAD = BCD + \left(180^{\circ} - \angle BCD\right) = 180^{\circ}, \]thus $AFED$ is cyclic. So, $\angle AFD = \angle AED$, so $\angle BFD = \angle AEC$. Thus, $\triangle BFD \sim \triangle AEC$. Since $AE = EC$ as $E$ is on the perpendicular bisector of $AC$, then $FB = FD$, as desired.

Claim:$F$ is the circumcenter of $\Delta BED$. Proof: See the diagram attached below. Let $\angle ABD=\angle ACD=\alpha$ and $\angle BCA=\angle BAC=\theta$. As $BE \perp AC$ we have, $\angle BEC=90-\alpha$. Further angle chasing yields, $\angle BXC=\alpha+\theta$ and $\angle XBP=90-\alpha-\theta$. As $EF \parallel BC$, $\angle DEF=\angle DCB=\alpha+\theta$. Also, $\angle BEC=90-\alpha \implies \angle BEF=180-(\alpha+\theta+90-\alpha)=90-\theta=\angle FBE \implies FB=FE$. Also, $\angle EDB=180-(\angle BCD+\angle DBC)=180-(\alpha+\theta+180-\alpha-2\theta)=\theta$. Now, $\angle EFB=180-((90-\theta)+(90-\alpha-\theta+\alpha))=2\theta$. Thus $\angle EFB=2 \angle EDB$ which implies, $F$ is the circumcenter of $\Delta BDE$ as desired. $\blacksquare$

We proceed by Pascal's theorem: Let $X=AE \cap (ABC), Y=ED \cap (ABC)$. By Pascal's theorem on $BAXYDC$, we have: $E, F, XY \cap BC$ to be collinear. Since $EF \parallel BC$, we have $XY \parallel BC$. Now, applying Pascal's theorem on $YACBDX$, we have: $AC \cap XD, XY \cap BC, AY \cap BD$ to collinear. Since $AC \parallel XD$, we must have $AY \parallel BD$ and therefore $AYCD$ is isosceles trapezium and we are done.

BRUH so easy but cute angle chase Observe that $\odot(ADEF)$ is cyclic due to Reim's theorem and we have $$\angle FBD=\angle ACE=\angle EAC=\angle FAE-\angle FAC=\angle FDE-\angle BDC=\angle FDB\implies FB=FD $$[asy][asy]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 = -9.009113082039914, xmax = 5.554079822616402, ymin = -6.283725055432363, ymax = 7.197427937915731; /* image dimensions */ draw((-4,5)--(-4.6,-1.6)--(2,-1)--(-2.8445621419951905,5.5644998311475025)--cycle, linewidth(0.7)); /* draw figures */ draw(circle((-1.55,1.45), 4.31335136523794), linewidth(0.7)); draw((-4,5)--(-4.6,-1.6), linewidth(0.7)); draw((-4.6,-1.6)--(2,-1), linewidth(0.7)); draw((2,-1)--(-2.8445621419951905,5.5644998311475025), linewidth(0.7)); draw((-2.8445621419951905,5.5644998311475025)--(-4,5), linewidth(0.7)); draw((-4,5)--(2,-1), linewidth(0.7)); draw((-2.8445621419951905,5.5644998311475025)--(-4.6,-1.6), linewidth(0.7)); draw(circle((-2.507415390026634,3.409671068854858), 2.1810444578610944), linewidth(0.7)); draw((-4.262312026764315,2.114567705592539)--(-2.8445621419951905,5.5644998311475025), linewidth(0.7)); draw((-4.6,-1.6)--(-0.5477443211717749,2.452255678828225), linewidth(0.7)); draw((-4.262312026764315,2.114567705592539)--(-0.5477443211717749,2.452255678828225), linewidth(0.7)); draw((-4,5)--(-0.5477443211717749,2.452255678828225), linewidth(0.7)); /* dots and labels */ dot((-4,5),linewidth(3pt) + dotstyle); label("$A$", (-4.3616629711751695,5.13977827050997), NE * labelscalefactor); dot((-4.6,-1.6),linewidth(3pt) + dotstyle); label("$B$", (-4.858337028824837,-2.0619955654101965), NE * labelscalefactor); dot((2,-1),linewidth(3pt) + dotstyle); label("$C$", (2.2015299334811482,-1.2992461197339227), NE * labelscalefactor); dot((-2.8445621419951905,5.5644998311475025),linewidth(3pt) + dotstyle); label("$D$", (-2.9071175166297154,5.831574279379148), NE * labelscalefactor); dot((-0.5477443211717749,2.452255678828225),linewidth(3pt) + dotstyle); label("$E$", (-0.3350554323725098,2.372594235033256), NE * labelscalefactor); dot((-4.262312026764315,2.114567705592539),linewidth(3pt) + dotstyle); label("$F$", (-4.734168514412421,1.8227050997782677), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

KaiSuii wrote: IDK if this is right place to post this problem, but here it is: The real root of the equation $8x^3-3x^2-3x-1=0$ can be written in the form $\frac{\sqrt[3]{a}+\sqrt[3]{b}+1}{c}$, where $a$, $b$, and $c$ are positive integers. Find $a+b+c$. try making a new topic (the button in the top right corner) for this problem, try using difference of cubes: the expression is equal to 9x^3 - (x+1)^3 which is factorable

Very messy sol

Notice that $\angle BFE = \angle ADE= 180 - \angle FBC$ so $(AFED)$ is cyclic. $FB=FD$ can be rewritten as $\angle AFD = 2 \angle FBD$ but $\angle FBD = \angle ACD$ so we need $\angle AFD = \angle AED = 2 \angle ACD$. But this is just true because $AE = EC$ from our isosclees condition so we are done.