hey,i got the soln to the geometry problem.in the triangle reflect M across AC.(that's the hint)well,i'm in a hurry.so i'll post the full soln later.for the time being here's the final equation that i got 2cos10cos(60+x)=sinx where x= angle MCB.evidently x=20 is a soln.i havent bothered to find the others(if any).therefore it turns out AB=BC. Regards vineet

yeah ... i know we aren't suppose to know this, but ... consider the function f(x)=2 cos10 cos60+x- sin x f'(x)=-2cos 10 sin 60+x - cos x and because x can be at most 80, we get that f'(x)<0 for x \in (0,80). Thus it is one-to-one, so it can only have one solution ... that is the one you found vineet. waiting for the rest of the solution ...

well,the soln i 'll give here is different from my original solution as it doesnt need the reflection tool.apply "trig ceva" with angle MCB as "x" we get sin20.sinx.sin40 = sin10.sin(80-x).sinsin30 which is equivalent to sin (80-x) = 4sinx.sin40.cos10 now the identity 2sina.cosb=sin(a-b) + sin(a+b) yields sin(80-x) = 2sinx.(sin 30 + sin50) which implies sin(80-x) = sinx +2sinxcos40 therefore 2sinxcos40 = 2sin(40-x)cos 40. rest is obvious. my previous soln was a hybrid of pure geometry and trigonometry (more of the latter).actually i got a third soln without trigonometry.if neone's willing to know abt it,i 'll post it. regards vineeet

yes vineet i'm quite curious ... i haven't gave this problem much tought, but anyway i'd like to see a purely geometric solution ( i'm in for the nicer solutions .. although generally i give ugly ones )

ya here is the solution using pure geo. reflect the point A across BM and name the point D.than triangle AMD is isosceles with vertex angle 2 (180 - <BMA) = 2(<MAB + <ABM) = 60. and hence is equilateral.also <DBA = 2<MBA = 40 and since <BAC = 50 we have DB perpendicular to AC.let E be the intersection of DB with CM.then <MED = 180-<CED=180 - (90-<ACE) = 90 +30 =120.therefore<MED +<DAM = 180.therfore quad. AMED is cyclic.therfore <DEA =<DMA = 60.we note that <DEC=<DEA=60.since AC perpendicular to DE theforewe see that A and C are symmetric across the line DE which implies BA=BC. i hope there iz nothing wrong with this proof bye

ups ... i was going through some of my materials today and i happened to find this problem as being question no.4 on the 25th USAMO, held on May 2nd, 1996

\[ \mathrm {A\ happy\ New\ Year,\ Mathlinks\ !} \]

we put : x=<MCB tHEN : <MBC=80-x then we use SEVA theorem and write this: (sin<MAB/sin<MAC).(sin<MCA/sin<MCB).(sin<MBC/sin<MBA)=1 ====> (sin10/sin40).(sin30/sinx).(sin(80-x)/sin20)=(sin(80-x))/4sin40cos10sinx therefor we understand this: sin(80-x)=2sinx.(2sin40cos10)=2sinx.(sin50+sin30)=sinx.(2cos40+1) ====>so: 2sinx.cos40=sin(80-x)-sinx=2sin(40-x)cos40 at the end we understand that :sinx=sin(40-x)====> x=40-x ====> <ACB=30+x=50 ===> x=20 ===><ACB=<BAC=50 THEN WE UNDERSTAND THAT THIS triangle is isosceles. sorry i dont know how i use LATEX

A solution with a 18 edges regular polygon without trigonometry . Image not found Domi Sorry for my poor english

Solution with an incenter: Reflection of AB in AM cuts CM at D. Triangle ADC is D-isosceles with <DAC = <ACD = 30°. Perpendicular bisector of AC through D cuts AB at B'. AM, DM bisect <DAB=20°, <B'DA = 120° => B'M bisects <AB'D = 40° => <MB'A = 20° => B' is identical with B => triangle ABC is B-isosceles.

There is a definite connection of this problem with Virgil's slicing problem III (What happened to I and II though ?) http://www.mathlinks.ro/Forum/viewtopic.php?t=179544 Both problems deal with triangles having internal angles 50,60,70. It seems that arith. prog. of angles is the key to solution. Thanks. M.T.

Take D the circumcenter of triangle AMC, see AMD equilateral, AD and BM perpendicular, there4 D is the reflection of A across BM, giving <ABD=40 deg, which makes BD and AC perpendicular, then remember AD=DC (D-circumcenter), this makes BD perpendicular bisector of AC and AB=BC. Best regards, sunken rock

That's an awesome solution, Domi, with the 18 sided regular polygon. How did you figure it out? That is a good method to create hard problems too. I just found this website which discusses it: http://www.cut-the-knot.org/triangle/TrigCeva.shtml. Using the applet on the website, here is another problem I created in case anyone is interested: In triangle ABC, angle ACB=30, angle ABC=110, and M is a point outside the triangle. Angle ACM=30 and angle MBC= 40. Find angle MAC. (You can use the applet to solve it).

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1996Problem5 Vo Duc Dien

vineet wrote: ya here is the solution using pure geo. reflect the point A across BM and name the point D.than triangle AMD is isosceles with vertex angle 2 (180 - <BMA) = 2(<MAB + <ABM) = 60. and hence is equilateral.also <DBA = 2<MBA = 40 and since <BAC = 50 we have DB perpendicular to AC.let E be the intersection of DB with CM.then <MED = 180-<CED=180 - (90-<ACE) = 90 +30 =120.therefore<MED +<DAM = 180.therfore quad. AMED is cyclic.therfore <DEA =<DMA = 60.we note that <DEC=<DEA=60.since AC perpendicular to DE theforewe see that A and C are symmetric across the line DE which implies BA=BC. i hope there iz nothing wrong with this proof bye how to get the idea of reflection?

Isn't it too easy for USAMO.....Just sine rule gives $\frac{BM}{CM},\frac{CM}{AM},\frac{AM}{BM}$ so multiplying them one obtains $\frac{sin(100-B)}{sin(B-20)}\frac{sin40}{sin30}\frac{sin20}{sin10}=1$ which is easy too solve....

I guess it was a tough problem if use of trignometry had been abandoned.. But here is a neat pure geo solution . The motivation lies in the fact that $\angle{BMC}=100=90+10$ something similar to the propert yof incentre,right?.So let us take a point $D$ on $MC$ such that $\angle{MAD}=10$. Observe that $M$ is the incentre of $ABD$.So $\angle{ADB}=120$. Also it can be easily observed that $BDA$ is congruent to $BDC$ So we get $AB=BC$.Done!

I have a solution using barycentric coordinates.... First represent M using Conways notation in terms of $A$ and $B$, then again using conways notation we get {$measuredangle{ACM}$/ which finishes the problem right away using sine rule

pinkpig wrote: ThinkThink wrote: Isn't this the oldest thread in AoPS? The oldest thread on AoPS got deleted. This is the second oldest. Really? What was the oldest thread about?

veranikaia wrote: I don't know. It could be.. it took place long before I was born. lol same

Nice problem(third USAMO geo ive solved lol) Let $\angle{MBC}=\theta,\angle{MCB}=(80-\theta)$, every angle here is in degree. Using trig Ceva, we can get the equation that $\frac{\sin20}{\sin40}*\frac{\sin10}{\sin20}*\frac{\sin\theta}{\sin(80-\theta)}=1$. Now we look at three different cases, $AB=AC;AC=BC;AB=BC$ $(1)AC=AB$, $\theta=45$ and this cant work. $(2)AC=BC$, $\theta=30,\angle{MCB}=50$. this cant work as well since $\frac{\sin10}{\sin20}*\frac{\sin50}{\sin20}\neq 1$ $(3),AB=BC,\theta=60$, we have to prove that $4\sin{20} \sin{20} \sin{80}=\sin{60}$ Write that $4\sin{(30-10)}\sin{(30+10)}\cos{10}=4(\frac{\cos^2{10}}{4}-\frac{3\sin^2{10}}{4})\cos{10}=4\cos^3{10}-3\cos{10}=\cos{30}=\sin{60}$ as desired

My solution using a Trigonometric Trick: A Trigonometric Trick: The relation $x+y = a+b < \pi $ is given, provided that $x$, $y$, $a$, $b$ are positive angle measures. If $\dfrac{\sin x}{\sin y} = \dfrac{\sin a}{\sin b}$ then, $x=a$ and $y=b$. (Since it is easy to prove, I will not write it here. I can add the proof if requested.) Let $\angle MBC = x$, $\angle MCB = y$. $x+y = 80^\circ$. By trigonometric Ceva, $$ \sin x \cdot \sin 30^\circ \cdot \sin 10^\circ = \sin y \cdot \sin 40^\circ \cdot \sin 20^\circ .$$Easily we can prove that $$ \sin 20^\circ \cdot \sin 40^\circ \sin 80^\circ = \dfrac{\sqrt{3}}{8} $$ Hence, $$ \dfrac{\sin x}{\sin y} = \dfrac{\sin 60^\circ}{\sin 20^\circ} $$ By the trigonometric trick, $x=60^\circ$ and $y=20^\circ$. Thus, the triangle $ABC$ is isosceles. (Lokman GÖKÇE)

ThinkThink wrote: veranikaia wrote: I don't know. It could be.. it took place long before I was born. lol same we don't know. the first post might be the welcome to aops or sth, but it's deleted This is the oldest thread still on aops today

Lol, looks like the mindset of constructing a bunch of points to make random isosceles triangles worked. Why are there so many trig solutions By angle chasing, the following constructions are possible: $P$ on $\overline{AB}$ satisfying $AP=PM=MB$ $Q$ on $\overline{AC}$ such that $\triangle MPQ$ is equilateral $R$ on $\overline{CM}$ such that $\triangle MQR$ is equilateral $S=\overline{AC} \cap \overline{BR}$ such that $QR=QS$ and $SC=SR$. Since $\angle MBR=\angle MRB=\angle CRS=\angle RCS$, $BMSC$ is cyclic. Thus, $\angle BCM=\angle BSM=\frac{\angle MQR}{2}=30^\circ$, so $\angle ACB=70^\circ$ and $\angle ABC=70^\circ$, as desired.

CyclicISLscelesTrapezoid wrote: Why are there so many trig solutions The AMSP book has a trig solution, and most people doing olympiad problems have the AMSP books.

wow oldest post ever cool

Let $\angle MBC=x$, then $\angle MCB=80^\circ-x$. By Trig Ceva: $$\frac{\sin\angle MBA}{\sin\angle MBC}\cdot\frac{\sin\angle MCB}{\sin\angle MCA}\cdot\frac{\sin\angle MAC}{\sin\angle MAB}=\frac{\sin20^\circ}{\sin x}\cdot\frac{\sin(80^\circ-x)}{\sin30^\circ}\cdot\frac{\sin40^\circ}{\sin10^\circ}=1.$$Let $f(x)=\frac{\sin20^\circ}{\sin x}\cdot\frac{\sin(80^\circ-x)}{\sin30^\circ}\cdot\frac{\sin40^\circ}{\sin10^\circ}$, we notice that $f'(x)>0$ so at most one value of $x$ works. But since a triangle exists with $x=60^\circ$, the triangle is isosceles with $\angle BCA=\angle BAC$.

Valentin Vornicu wrote: Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles. was this the first post on aops? url is https://artofproblemsolving.com/community/c6h2 https://artofproblemsolving.com/community/c6h1 doesn't exist

Nice question! (Trig functions in degrees) WLOG let $AB=1,$ angle chasing gives $\angle AMB=150,$ and $\angle AMC=110,$ using law of sines on $AMB,$ and $AMC,$ yields $AB\cdot \frac{\sin(20)}{\sin(150)}=AM=2\sin(20),$ now $AC=AM\cdot \frac{\sin(110)}{\sin(30)},$ note that $\sin(110)=\sin(90+20)=\cos(20),$ so this is $AM\cdot 2\cos(20)=4\sin(20)\cos(20)=2\sin(40),$ using law of cosines in $ABC,$ gives $BC^2=1^2+(2\sin(40))^2-4\sin(40)\cos(50)=1,$ hence $AB=BC,$ and triangle $ABC,$ is isosceles $\blacksquare$

Let $\angle MBC=\alpha$ and $\angle MCB=\beta$. Then $\alpha +\beta =80^\circ =60^\circ +20^\circ$, so it suffices to prove that $\frac{\sin \alpha}{\sin \beta}=\frac{\sin 60^\circ}{\sin 20^\circ}$. From the Trigonometric form of Ceva's Theorem we have $\frac{\sin 10^\circ}{\sin 40^\circ}\frac{\sin 30^\circ}{\sin 20^\circ}\frac{\sin \alpha}{\sin \beta}=1$, so it suffices to prove that $\frac{\sin 10^\circ}{\sin 40^\circ}\frac{\sin 30^\circ}{\sin 20^\circ}\frac{\sin 60^\circ}{\sin 20^\circ}=1$. Now using the identities $\sin x\sin y=\frac{\cos (x-y)-\cos (x+y)}{2}$ and $\cos x\sin y=\frac{\sin (x+y)-\sin (x-y)}{2}$ we obtain\begin{align*}\sin 10^\circ \sin 30^\circ & =\frac{\cos 20^\circ -\cos 40^\circ}{2}, \\ \sin 40^\circ \sin 20^\circ & =\frac{\cos 20^\circ -\cos 60^\circ}{2}, \\ \cos 20^\circ \sin 60^\circ & =\frac{\sin 80^\circ +\sin 40^\circ}{2}, \\ \cos 40^\circ \sin 60^\circ & =\frac{\sin 100^\circ +\sin 20^\circ}{2}=\frac{\sin 80^\circ +\sin 20^\circ}{2}, \\ \cos 20^\circ \sin 20^\circ & =\frac{\sin 40^\circ}{2}, \\ \cos 60^\circ \sin 20^\circ & =\frac{\sin 80^\circ -\sin 40^\circ}{2}. \end{align*}Hence\begin{align*}\frac{\sin 10^\circ}{\sin 40^\circ}\frac{\sin 30^\circ}{\sin 20^\circ}\frac{\sin 60^\circ}{\sin 20^\circ} & =\frac{\cos 20^\circ -\cos 40^\circ}{\cos 20^\circ -\cos 60^\circ}\frac{\sin 60^\circ}{\sin 20^\circ} \\ & =\frac{\cos 20^\circ \sin 60^\circ -\cos 40^\circ \sin 60^\circ}{\cos 20^\circ \sin 20^\circ -\cos 60^\circ \sin 20^\circ} \\ & =\frac{(\sin 80^\circ +\sin 40^\circ )-(\sin 80^\circ +\sin 20^\circ )}{(\sin 40^\circ )-(\sin 80^\circ -\sin 40^\circ )} \\ & =\frac{\sin 40^\circ -\sin 20^\circ}{2\sin 40^\circ -\sin 80^\circ}. \end{align*}Therefore it suffices to prove that $\sin 40^\circ -\sin 20^\circ =2\sin 40^\circ -\sin 80^\circ$, i.e. that $\sin 80^\circ =\sin 40^\circ +\sin 20^\circ$. We have\begin{align*}\sin 40^\circ +\sin 20^\circ & =2\sin \frac{40^\circ +20^\circ}{2}\cos \frac{40^\circ -20^\circ}{2} \\ & =2\sin 30^\circ \cos 10^\circ \\ & =\cos 10^\circ \\ & =\sin 80^\circ , \end{align*}as wanted.

Valentin Vornicu wrote: Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles. o7 WLOG, let $AB=1$. By LoS, $AM=AB\frac{\sin20^\circ}{\sin150^\circ}=2\sin20^\circ$. By LoS, $AC=\frac{AM}{\sin30^\circ}\sin110^\circ=4\sin20^\circ\sin110^\circ=2\sin40^\circ$. By LoC, $BC^2=AB^2+AC^2-2ABAC\cos50^\circ=1+4\sin^240^\circ-4\sin40^\circ\cos50^\circ=1$. Therefore, $AB=BC=1$, as desired.

RyanWang wrote: Valentin Vornicu wrote: Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles. was this the first post on aops? url is https://artofproblemsolving.com/community/c6h2 https://artofproblemsolving.com/community/c6h1 doesn't exist yes, I think so

Here is a pure geometric approach: