Let $ABC$ be a triangle with $\angle{BAC} > 90$. Let $D$ be a point on the segment $BC$ and $E$ be a point on line $AD$ such that $AB$ is tangent to the circumcircle of triangle $ACD$ at $A$ and $BE$ is perpendicular to $AD$. Given that $CA=CD$ and $AE=CE$. Determine $\angle{BCA}$ in degrees.
Problem
Source:
Tags: geometry, gguu, trigonometry
20.01.2019 15:21
i got 60 as answer
20.01.2019 15:25
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}}$
20.01.2019 16:36
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?
20.01.2019 16:46
@above, since, $BE \perp AX $ and we already have $AB=BX \implies \Delta ABX$ is isosceles, hence, $AE=EX$
20.01.2019 16:53
Bash bash bash Simple trig
20.01.2019 16:53
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?
20.01.2019 17:01
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?
20.01.2019 17:08
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, $3x - 180 = 90 - x$ . $\implies \angle ACB = 180 - 2x = 45$
20.01.2019 17:09
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?
20.01.2019 17:12
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.
20.01.2019 17:15
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 $cos(2x)\cdot cos(x) = -cos(2x)$
20.01.2019 17:39
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} $.
20.01.2019 19:28
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...
20.01.2019 19:41
^ Evaluators are not so compassionate in giving partials. As TDP said the point slabs appear to be 0,1,12,17.
20.01.2019 20:00
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 $AC$.
20.01.2019 20:02
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
20.01.2019 20:11
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 ??
21.01.2019 10:40
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
21.01.2019 13:18
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
31.01.2021 18:19
Lightprince wrote: i got 60 as answer Then you are wrong because then..... angle B would be 0 degree.....
09.02.2021 11:49
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 $\angle BCA=45^\circ$ $\blacksquare$
30.04.2021 19:16
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 $AC \implies \angle EGA = 90^\circ$ $\implies 2x + 2x = 90^\circ \implies 2x = 45^\circ = \angle BCA$
Attachments:

19.08.2021 20:25
Hi there! I just made a video about it. Hope you like it! https://youtu.be/UVFx1gVWP4w
19.08.2021 20:42
nabodorbuco wrote: Hi there! I just made a video about it. Hope you like it! https://youtu.be/UVFx1gVWP4w Very well explained!!
05.03.2022 16:25
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.
08.03.2023 04:06
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 $$\frac{\sin\angle BAE}{\sin\angle CAE}=\frac{\sin\angle ABE}{\sin\angle ACE}\cdot\frac{\sin\angle BCE}{\sin\angle CBE} \leftrightarrow \sin(2x)\sin(x)=\sin(90-2x)\sin(90-3x)\leftrightarrow x=22.5^{\circ} \rightarrow 2x=\boxed{45^{\circ}}.$$
30.06.2023 16:18
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 $\angle{BCA}=180^{\circ}-2\theta=\boxed{45^{\circ}}$ $\blacksquare$
20.07.2023 04:04
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, $180-2\alpha = \boxed{45^{\circ}}$.
31.10.2023 05:16
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 \[\cos\alpha=\cos\angle FDE=\frac{a/2}{DE}=\frac{1}{\sqrt{4+2\sqrt2}}\Longrightarrow \cos(2\alpha)=\frac{2}{4+2\sqrt2}-1=\frac{1}{2+\sqrt2}-1=\frac{2-\sqrt2}{2}-1=\frac{-\sqrt2}{2}\]\[\Longrightarrow \cos(\angle ACB)=\frac{\sqrt2}{2}\Longrightarrow \boxed{\angle ACB=45^{\circ}}.\]
10.01.2024 11:26
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, $AE=CE \Rightarrow \Delta AEF \cong \Delta CEF$. $\Rightarrow \angle AFE= 90^{\circ} \Rightarrow \angle FAX+ \angle FXA= 90^{\circ} \Rightarrow 2(90^{\circ}-2\theta)=90^{\circ}$ $\Rightarrow 2\theta=45^{\circ} \Rightarrow \angle BCA= \boxed{45^{\circ}}$
Attachments:

15.03.2024 00:06
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.
24.08.2024 05:00
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.
21.11.2024 14:51