i got 60 as answer

My solution (motivated from Tumon2001's hint!): Let $\angle BCA=x$ and Let $AE \cap \odot (ABC) =X$, then, $\angle BXA=\angle BCA=x=\angle BAX$ $ \implies AB=BX \text{ and } AE=EX=EC \implies (E) \equiv \odot (ABC) \implies \angle ABX=180^{\circ}-2x=90^{\circ} \implies \boxed{x= 45^{\circ}}$

AlastorMoody wrote: My solution (motivated from Tumon2001's hint!): Let $\angle BCA=x$ and Let $AE \cap \odot (ABC) =X$, then, $\angle BXA=\angle BCA=x=\angle BAX$ $ \implies AB=BX \text{ and } AE=EX=EC \implies (E) \equiv \odot (ABC) \implies \angle ABX=180^{\circ}-2x=90^{\circ} \implies \boxed{x= 45^{\circ}}$ AE=EX?

@above, since, $BE \perp AX $ and we already have $AB=BX \implies \Delta ABX$ is isosceles, hence, $AE=EX$

Bash bash bash Simple trig

AlastorMoody wrote: @above, since, $BE \perp AX $ and we already have $AB=BX \implies \Delta ABX$ is isosceles, hence, $AE=EX$ Ok, missed that. One doubt - I computed all the angles in the figure without construction, but could not reach the answer. How many would I get for this?

If you do by sign rule, and get the correct trignometric equation in angle C, and you leave the equation as it is, how much can you get?

Here's my solution(SYNTHETIC ): Let $\angle ADC = x$. After some angle chase we get, $\angle ABC = \angle BCE = 3x - 180.$. So, $AB \parallel CE$. Now extend $BA$ to $X$ such that $XC \parallel AE$. Notice that $AECX$ is a rhombus. Therefore $\angle EXC = 90 - x = \angle BEC$. Therefore, $BEXC$ is cyclic. $\implies \angle XBC = \angle XEC.$ Therefore, $3x - 180 = 90 - x$ . $\implies \angle ACB = 180 - 2x = 45$

JRyno wrote: If you do by sign rule, and get the correct trignometric equation in angle C, and you leave the equation as it is, how much can you get? Dunno, maybe 14-15 in the worst case?

MathPassionForever wrote: JRyno wrote: If you do by sign rule, and get the correct trignometric equation in angle C, and you leave the equation as it is, how much can you get? Dunno, maybe 14-15 in the worst case? Wtf 2-3 in the best case I got a biquadratic equation in cos phi before 40 seconds the exam ended and then rage quotes.

Kayak wrote: MathPassionForever wrote: JRyno wrote: If you do by sign rule, and get the correct trignometric equation in angle C, and you leave the equation as it is, how much can you get? Dunno, maybe 14-15 in the worst case? Wtf 2-3 in the best case I got a biquadratic equation in cos phi before 40 seconds the exam ended and then rage quotes. My friend PMed me the one he got, it was $cos(2x)\cdot cos(x) = -cos(2x)$

Looks like someone has already posted my solution above (post #3) .....to reduce it even further... Let $AE\cap \odot (ABC)=F $. Angle chasing gives $BA=BF\implies EA=EF $. Since, $EA=EC $, so $E $ is the center of $\odot (ABC)$. Thus, $\angle ACB=\frac {1}{2}\angle AEB=45^{\circ} $.

Kayak wrote: MathPassionForever wrote: JRyno wrote: If you do by sign rule, and get the correct trignometric equation in angle C, and you leave the equation as it is, how much can you get? Dunno, maybe 14-15 in the worst case? Wtf 2-3 in the best case I got a biquadratic equation in cos phi before 40 seconds the exam ended and then rage quotes. Really? Surely you would get more? Assuming it can be simplified in just 1-2 more steps...

^ Evaluators are not so compassionate in giving partials. As TDP said the point slabs appear to be 0,1,12,17.

what does INMO stand for ? Answer is $45$. Take $K$ on $BC$ such that $AK$ is perpendicular to $BC$ and prove that $EK$ passes through the midpoint of $AC$.

mela_20-15 wrote: what does INMO stand for ? Indian National MAthematical Olympiad. The second (or third) stage of MOP in India v Mathematical Olympiad Program I think

monsterDJ wrote: mela_20-15 wrote: what does INMO stand for ? Indian National MAthematical Olympiad. The second (or third) stage of MOP in India MOP ??

AlastorMoody wrote: monsterDJ wrote: mela_20-15 wrote: what does INMO stand for ? Indian National MAthematical Olympiad. The second (or third) stage of MOP in India MOP ?? Mathematics Olympiad Programme

TheDarkPrince wrote: Bash bash bash Simple trig JRyno wrote: If you do by sign rule, and get the correct trignometric equation in angle C, and you leave the equation as it is, how much can you get? I did it for you .Pre boards are such a horrible bore. @below Well I am a noob at these stuffs so doesnt matter

Lightprince wrote: i got 60 as answer Then you are wrong because then..... angle B would be 0 degree.....

Another Solution can be.. First observe By Tangent-Secant theorem we have $\angle BAE=\angle BCA$ Now we will choose a variable Point $B^*\in BE$ Such $EB^*=AD=AC$ Now there are $2$ cases Case 1-: if $B^*E<BE$ Then observe $\angle B^*CA=\angle B^*AE=45^\circ \implies$ $\angle BCA+\angle BCB^*=\angle BAE -\angle BAB^*$ Which is only possible if $\angle BCB^*=\angle BAB^*=0^\circ$ or $B^*\equiv B$ Hence $\angle BCA=45^\circ$ in this case. Case 2-: if $B^*E>BE$ Then $\angle B^*AE=\angle BAE +\angle B^*AB=\angle BCA-\angle B^*CB$ And here also it is only possible if $\angle B^*AB=\angle B^*CB=0^\circ$ Hence $B^*\equiv B$ and $\angle BCA=45^\circ$ $\blacksquare$

Simple solution: Let $\angle ADC = 2x$ and $F$ be the foot of perpendicular from $A$ to $BC$ and $G = EX \cap AC$. $\implies ABEF$ is cyclic. Simple angle chase done in picture $\angle GFA = \angle GAF \implies AG=GF$ and $GF = GC \implies G$ is midpoint of $AC \implies \angle EGA = 90^\circ$ $\implies 2x + 2x = 90^\circ \implies 2x = 45^\circ = \angle BCA$

Hi there! I just made a video about it. Hope you like it! https://youtu.be/UVFx1gVWP4w

nabodorbuco wrote: Hi there! I just made a video about it. Hope you like it! https://youtu.be/UVFx1gVWP4w Very well explained!!

we have $\angle ECA = \angle EAC = \angle ADC = x$. $\angle BAE = \angle 180 - 2x = \angle AEC$ so $CE || AB$. Let line from $C$ parallel with $AE$ meet $AB$ at $S$. we have $\angle CSE = \angle 90 - x = \angle CBE$ so $BECS$ is cyclic so $\angle 180 = \angle EBS + \angle ECS = 2x - 90 + 2x$ so $4x = 270$ so $x = 67.5$. $\angle BCA = \angle DCA = \angle 180 - 2x = 45$. we're Done.

Let $\angle BCA = 2x$. Notice that by angle chasing, we can directly get the measure of all angles in the problem statement: \begin{align*} \angle CAE = \angle ADC = \angle BDE &= 90-x \\ \angle BCA = \angle BAD &= 2x \\ \angle ABD = \angle BCE &= 90-3x \\ \angle CBE &= x \\ \angle ADB = \angle CDE &= 90+x. \\ \end{align*} Now, by Quadrilateral Ratio Lemma, we have $$\frac{\sin\angle BAE}{\sin\angle CAE}=\frac{\sin\angle ABE}{\sin\angle ACE}\cdot\frac{\sin\angle BCE}{\sin\angle CBE} \leftrightarrow \sin(2x)\sin(x)=\sin(90-2x)\sin(90-3x)\leftrightarrow x=22.5^{\circ} \rightarrow 2x=\boxed{45^{\circ}}.$$

We extend $BA$ to point $P$ such that $CP \parallel AE$ this gives $AECP$ to be a rhombus denote $\angle{ADC}=\theta$ a, since $BE \perp AD$ we have $\angle{DBE}=90^{\circ}-\theta$ , also we have $\angle{ADC}=\angle{DAC}=\angle{ECA}=\theta$ we have $\angle{DCA}=180^{\circ}-2\theta \implies \angle{ECD}=3\theta-180^{\circ}$ since $AECP$ is a rhombus we have $\angle{EPC}=90^{\circ}-\theta \implies BECP$ is a cyclic quadrilateral since we have $\angle{DBE}=90^{\circ}-\theta$ so we have $\angle{ECD}=\angle{APE} \implies 3\theta-180^{\circ}=90^{\circ}-\theta \implies 180-2\theta=45^{\circ}$ so $\angle{BCA}=180^{\circ}-2\theta=\boxed{45^{\circ}}$ $\blacksquare$

Bruh Let $\alpha = \angle{DAC}$. It suffices to find $180-2\alpha$. Note that $\angle{AEC} = 180-2\alpha = \angle{BAE}$, because of tangent properties. Define $X$ as the foot of the altitude from $C$ onto $AD$. Note that $\triangle{ABE}\sim \triangle{ECX}$ and also $\triangle{CXD} \sim \triangle{BED}$. Because of this, we have \[\frac{XD}{DE} = \frac{XE}{AE}\implies \frac{AD}{2\cdot DE} = \frac{AD+2\cdot DE}{2\cdot (AD+DE)} \implies AD = \sqrt{2}\cdot DE\]and we also have \[\frac{AB}{AE}=\frac{AB}{CE} = \frac{CX}{BE} = \frac{XD}{DE} \implies AB = \sqrt{2}AE\]Thus, $180-2\alpha = \boxed{45^{\circ}}$.

Let $F$ be the other intersection of $(ACD)$ and $CE$ and let $\alpha=\angle ADC$. Then \[\alpha=\angle ADC=\angle DAC=\angle ACE,\]so \[\angle BAC+\angle ACE=\angle BAD+2\alpha=\angle ACD+\angle DAC+\angle ADC=180^{\circ},\]so $AB\parallel CE$ and $AC\parallel DF$. Let $P$ be the foot of the perpendicular from $C$ to $AD$. Note that $AP=PD$. We have $\triangle BED\sim\triangle CPD$ by AA similarity. Now consider the negative homothety $h$ centered at $D$ such that $h(B)=C$. Then by the similarity, $h(E)=P$. We have $h(AB)=CE$ since they are parallel. Since $h(AE)=AE$, $h(A)=h(AB\cap AE)=CE\cap AE=E$. So $DE^2=(DA)(DP)=\frac12 DA^2$, so $\sqrt2 DE=DA$. Now let $DF=a$. Then $AC=(\sqrt2+1)a$, so $CD=AF=(\sqrt2+1)a$. If we let $AD=b$, applying Ptolemy's on cyclic quadrilateral $ADFC$ gives \[(\sqrt2+1)a^2+b^2=(\sqrt2+1)^2 a^2\Longrightarrow b=\sqrt{2+\sqrt2} a\Longrightarrow DE=\sqrt{4+2\sqrt2}\cdot\frac{a}{2}.\]Thus \[\cos\alpha=\cos\angle FDE=\frac{a/2}{DE}=\frac{1}{\sqrt{4+2\sqrt2}}\Longrightarrow \cos(2\alpha)=\frac{2}{4+2\sqrt2}-1=\frac{1}{2+\sqrt2}-1=\frac{2-\sqrt2}{2}-1=\frac{-\sqrt2}{2}\]\[\Longrightarrow \cos(\angle ACB)=\frac{\sqrt2}{2}\Longrightarrow \boxed{\angle ACB=45^{\circ}}.\]

Let $AX\perp BC$ such that $X$ lies on $BC$. Extending $EX$ to meet $AC$ at $F$. Let $\angle BCA= 2\theta \Rightarrow \angle CAD= \angle CDA= 90^{\circ}-\theta$. (As $\Delta CAD$ is isosceles.) Also, $AE=CE \Rightarrow \angle EAC= \angle ACE = 90^{\circ}-\theta \Rightarrow \angle AEC= 2\theta$. As $AB$ is tangent to the circumcircle of $\Delta CAD \Rightarrow \angle DAB = \angle ACD= 2\theta$. (by alternate segment theorem). From the previous results, $\angle AEC=\angle EAB = 2\theta \Rightarrow AB \parallel CE$. $\angle BEA = \angle BXA = 90^{\circ} \Rightarrow ABEX$ is cyclic. $\Rightarrow \angle BAE= \angle BXE= 2\theta = \angle CXF$ (vertically opposite). Thus, $\Delta CFX$ is isosceles $\Rightarrow XF=CF$. $\angle AXF= 90^{\circ} -\angle CXF= 90^{\circ}-2\theta = \angle XAF$. Thus, $\Delta AXF$ is isosceles $\Rightarrow AF=XF$. But $XF=CF\Rightarrow AF=CF$. Also, $AE=CE \Rightarrow \Delta AEF \cong \Delta CEF$. $\Rightarrow \angle AFE= 90^{\circ} \Rightarrow \angle FAX+ \angle FXA= 90^{\circ} \Rightarrow 2(90^{\circ}-2\theta)=90^{\circ}$ $\Rightarrow 2\theta=45^{\circ} \Rightarrow \angle BCA= \boxed{45^{\circ}}$

Let $\angle ACD=2a$. C1: I claim that $AB\parallel CE$. This is because since $AB$ is tangent to $(ADC)$ at $A$, we have that \[\angle BAE=\angle ACD=2a.\]Additionally, since $AE=CE$, we have that \[\angle AEC=180-2\angle EAC=180-2\angle DAC=\angle ACD=2a,\]since $CA=CD$. Therefore, since $\angle BAE=\angle AEC$, we have that $AB\parallel CE$, proving our claim. C2: I claim that $\triangle BDA\sim \triangle CDE$. This follows directly from (C1). Since $AB\parallel CE$, this gives us that \[\angle ABD=\angle ECD,\]and \[\angle BAD=\angle DEC,\]meaning that $\triangle BDA\sim \triangle CDE$, proving our claim. C3: Let $X\neq D$ be on line $AB$ so that $EX=DX$. I claim that $\triangle BDX\sim CDA$. Note that since $\angle BED=90$ and $ED=DX$, we have that $\triangle BDX$ is isosceles with $BD=BX$. Since $\triangle CDA$ is also isosceles with $CA=CD$, to prove that they're similar, we just need to prove that $\angle BDE=\angle CDA$. This is obvious since lines $AE$ and $BC$ intersect at $D$. Therefore $\triangle BDX\sim CDA$, proving our claim. C4: I claim that $\frac{AD}{DE}=\sqrt{2}$. This is because by (C2), we have that $\triangle BDA\sim \triangle CDE$, meaning that \[\frac{BD}{DC}=\frac{AD}{DE}.\]Additionally, by (C3), we also have that $\triangle BDA\sim \triangle CDE$, meaning that \[\frac{BD}{DC}=\frac{XD}{AD}=\frac{2*DE}{AD}.\]Equating the two equations gives that \[2*\frac{DE}{AD}=\frac{AD}{DE},\]implying that $\frac{AD}{DE}=\sqrt{2}$, proving our claim. C5: I claim that $\frac{EA}{AB}=\frac{\sqrt{2}}{2}$. This is because \[\frac{EA}{AB}=\frac{EC}{AB}=\frac{ED}{AD}=\frac{\sqrt{2}}{2},\]by (C4), proving our claim. Finally, notice that $\frac{\sqrt2}{2}\frac{EA}{AB}=\cos(2a)$, meaning that \[\angle BCA=\angle ACD=2a=45,\]finishing the problem.

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Sketch: Angle chase for $AC$ tangent to $(CDE)$. Invert at $A$ preserving $(CDE)$. Then we see that $D\leftrightarrow E$, $C$ is fixed, and that $B$ is sent to the foot of $D$ onto $AB$. Note that $ACEB'$ is cyclic. We reverse reconstruct $E$. Let $E$ be $AD\cap (ACB')$. Then $(CDE)$ is tangent to $AB$. Note that by angle chase, \[\measuredangle DAB'=\measuredangle BCA=\measuredangle AEC=\measuredangle AB'C\]Letting $CB'\cap AD=X$, we see by Thales that $X$ is the midpoint of $AD$. However, $CX$ is then a symmedian of $\triangle CAD$ (Isos triangle). This means that $B'D$ is tangent to $(ACD)$. Clearly, we can get $\angle BCA=45^\circ$ by alternate segment theorem.