Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?
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Problem
Source: IMO ShortList 2001, geometry problem 8
Tags: geometry, ratio, trigonometry, similar triangles, angle bisector, IMO, IMO 2001
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orl
30.09.2004 20:47
Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions
probability1.01
03.10.2005 14:54
Weird solution: We will have $\angle ABQ = \angle QBC = x$ Let R lie on AC s.t. RAB is equilateral. Let T lie on AB extended past B such that BT = BP, and let U lie on AC such that UAT is equilateral. Since AU = AT = AB + BP = AQ + QB, we have QB = QU. As a result, we calculate $\angle QBU = 90 - \dfrac{60 + \angle ABQ}{2} = 60 - \dfrac{x}{2}$. Meanwhile, $\angle QBR = 60 - x$, so we have $\angle RBU = \angle QBU - \angle QBR = \dfrac{x}{2}$. Then suppose the bisector of $\angle BTP$ intersects BR at X. Then since $\angle BRT = \angle RBU = \dfrac{x}{2} = \angle BTX$, we have similar triangles, and by equal ratios $BT^2 = BX \cdot BR$. Equivalently, $BP^2 = BX \cdot BR$, so triangles BXP and BPR are similar; in particular, BX = XP. Since TX is the bisector of $\angle BTP$, we have then that T, B, X, and P are concyclic. Then $\dfrac{x}{2} = \angle XTP = \angle XBP = \angle ABC - \angle ABR = 2x - 60$. Solving for x, x = 40. Then $\angle ABC = 80 \implies \angle ACB = 40$.
The QuattoMaster 6000
11.04.2008 20:12
Consider points $ X$ and $ Y$ on $ AB$ and $ AC$ respectively such that $ BY=BP$ and $ BQ=QX$. Also, let $ AP$ meet $ YX$ at $ Z$ and $ BX$ at $ W$. Assume that $ C\ne X$. Notice that $ AY=AB+BY=AB+BP=AQ+BQ=AQ+QX=AX$ and $ \angle XAY=60$, so $ \triangle AXY$ is equilateral. Now, let $ \angle ABQ=\alpha=\angle QBC$. Thus, $ \angle AQB=180-\angle BAC-\angle ABQ=120-\alpha$. Also, $ \angle QBX=\angle QXB=\frac{\angle AQB}{2}=60-\frac{\alpha}{2}$. Hence, $ \angle BXY=\angle AXY-\angle AXB=60-(60-\frac{\alpha}{2})=\frac{\alpha}{2}$. Now, since $ BP=BY$, we have that $ \angle BPY=\angle BYP=\frac{\angle ABC}{2}=\alpha$. Thus, $ \angle PYZ=\angle AYX-\angle AYP=60-\alpha$. Since $ AZ$ is a bisector of equilateral $ \triangle AXY$, we see that $ AZ$ is the perpindicular bisector of $ YX$, which implies that $ \angle PXY=\angle PYZ=60-\alpha\implies \angle PXW=\angle PXY-\angle WXY=60-\frac{3\alpha}{2}$. Notice that $ \angle CBX=\angle QBX-\angle QBC=60-\frac{3\alpha}{2}$. Therefore, $ \angle CBW=\angle PXW=\angle PYW$ because of the fact that $ PW$ is the perpindicular bisector of $ YX$. This means that $ PWYB$ is cyclic. Therefore, $ \angle XWZ=\angle ZWY=\angle PBY=180-\angle ABC=180-2\alpha$. Since $ \angle ZWX=90-\angle WXZ=90-\frac{\alpha}{2}$, we see that $ 90-\frac{\alpha}{2}=180-2\alpha$, so $ 90=\frac{3\alpha}{2}\implies \alpha=60$. This means that $ \angle ABC=2\alpha=120$, so $ \angle ACB=180-\angle BAC-\angle ABC=180-60-120=0$, which is clearly impossible. This means that we have a contradiction, so $ C=X$. Hence, $ QC=QB$, so $ \alpha=180-\angle BAC-\angle ABC=120-2\alpha\implies \alpha=40$. Therefore, $ \angle ABC=2\alpha=80$ and $ \angle ACB=\alpha=40$.
Virgil Nicula
11.04.2008 22:25
Let $ ABC$ be a triangle with $ A = 60^{\circ}$ . Let $ AP$ bisect $ \widehat { BAC}$ and let $ BQ$ bisect $ \widehat {ABC}$ , where $ P\in BC$ and $ Q\in AC$ . If $ AB + BP = AQ + QB$ , then what are the angles of the triangle ? Proof I (metric). Denote $ l_b = BQ$ . Thus, $ l_b = \frac {2ac}{a + c}\cdot\cos\frac B2$ . Therefore, $ AB + BP = AQ + QB$ $ \Longleftrightarrow$ $ c + \frac {ac}{b + c} = \frac {bc}{a + c} + l_b$ $ \Longleftrightarrow$ $ c(c + a)(c + b) + ac(a + c) = bc(b + c) + 2ac(b + c)\cdot\cos\frac B2$ $ \Longleftrightarrow$ $ ab + (a + c)^2 - b^2 = 2a(b + c)\cdot\cos\frac B2$ $ \Longleftrightarrow$ $ ab + 4p(p - b) = 2a(b + c)\cdot\cos\frac B2$ $ \Longleftrightarrow$ $ ab + 4ac\cdot\cos^2\frac B2 = 2a(b + c)\cdot \cos\frac B2$ $ \Longleftrightarrow$ $ b\left(1 - 2\cdot\cos\frac B2\right) = 2c\cdot\cos\frac B2\cdot\left(1 - 2\cdot\cos\frac B2\right)$ $ \Longleftrightarrow$ $ b = 2c\cdot\cos\frac B2$ $ \Longleftrightarrow$ $ \sin B = 2\cdot\sin C\cdot\cos\frac B2$ $ \Longleftrightarrow$ $ 2\sin\frac B2\cos\frac B2 = 2\sin C\cos\frac B2$ $ \Longleftrightarrow$ $ \sin\frac B2 = \sin C$ $ \Longleftrightarrow$ $ B = 2C$ $ \Longleftrightarrow$ $ C = 40^{\circ}$ , $ B = 80^{\circ}$ because $ A = 60^{\circ}$ . Proof II (trigonometric) Denote ${ x=m(\widehat ABQ})$ . I"ll apply the Sinus' theorem in the triangles $ ABP$ , $ ABQ$ : $ AB+BP=AQ+QB$ $ \Longleftrightarrow$ $ 1+\frac {BP}{BA}=\frac {AQ+QB}{AB}$ $ \Longleftrightarrow$ $ 1+\frac {\sin 30}{\sin (30+2x)}=\frac {\sin x+\sin 60}{\sin (60+x)}$ $ \Longleftrightarrow$ $ \sin (60+x)\left[\sin (30+2x)+\frac 12\right]=\sin (30+2x)(\sin x+\sin 60)$ $ \Longleftrightarrow$ $ \cos (x-30)-\cos (90+3x)+\cos (30-x)=$ $ \cos (x+30)-\cos (30+3x)+\cos (2x-30)-\cos (90+2x)$ $ \Longleftrightarrow$ $ \underline {\cos (x-30)}+\underline {\underline {\sin 3x}}+\underline {\underline {\underline {\cos (x-30)}}}=$ $ \underline {\underline {\underline {\cos (x+30)}}}-\underline {\underline {\cos (30+3x)}}+\cos (2x-30)+\underline {\sin 2x}$ $ \Longleftrightarrow$ $ [\cos (x-30)-\cos (90-2x)]+[\sin 3x+\sin (60-3x)]=$ $ [\cos (x+30)-\cos (x-30)]+\cos (2x-30)$ $ \Longleftrightarrow$ $ 2\sin \left(30-\frac x2\right)\sin\left(60-\frac {3x}{2}\right)+2\sin 30\cos (3x-30)=\cos (2x-30)-2\sin 30\sin x$ $ \Longleftrightarrow$ $ \cos (3x-30)+2\sin\left(30-\frac x2\right)\sin \left(60-\frac {3x}{2}\right)=\cos (2x-30)-\cos (90-x)$ $ \Longleftrightarrow$ $ \cos (3x-30)+2\sin\left(30-\frac x2\right)\sin \left(60-\frac {3x}{2}\right)=2\sin \left(30+\frac x2\right)\sin\left(60-\frac {3x}{2}\right)$ $ \Longleftrightarrow$ $ \sin (120-3x)=2\sin\left(60-\frac {3x}{2}\right)\left[\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)\right]$ $ \Longleftrightarrow$ $ 2\sin\left(60-\frac {3x}{2}\right)\cos\left(60-\frac {3x}{2}\right)=2\sin\left(60-\frac {3x}{2}\right)\left[\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)\right]$ $ \Longleftrightarrow$ $ \sin\left(60-\frac {3x}{2}\right)=0\ \ \vee\ \ \cos\left(60-\frac {3x}{2}\right)=\sin\left(30+\frac x2\right)-\sin\left(30-\frac x2\right)$ . Thus, $ \boxed {\ x=40\ }$ or $ \sin\left(30+\frac {3x}{2}\right)+\sin\left(30-\frac x2\right)=\sin\left(30+\frac x2\right)$ $ \Longleftrightarrow$ $ 2\sin\left(30+\frac x2\right)\cos x=\sin\left(30+\frac x2\right)$ $ \Longleftrightarrow$ $ x\in\emptyset$ .
vishalarul
09.07.2009 21:02
Interesting problem
[geogebra]3cd2588b3a8ec28b3adecf64c88d939dc79c0a23[/geogebra]
Define B' on ray AB such that BP = BB'.
Define C' on ray AC such that QB = QC'.
Let $ \alpha = \angle ABQ = \angle QBC$.
Then AB + BP = AQ + QB implies AB + BB' = AQ + QC', so that AB' = AC' with angle B'AC' = 60 degrees. Hence, $ \boxed{\triangle B'AC'\text{ is equilateral}}$.
Suppose that B, P, C' were not collinear.
By simple angle chasing,Since $ \angle ABC = 2 \alpha$, $ \boxed{\angle BB'P = \angle BPB' = \alpha}$. So this means that $ BQ || B'P$. Since AB'C' is equilateral, $ \boxed{\angle PB'C' = 60^{\circ} - \alpha}$. Since $ PB'C'$ is isosceles (P is on the angle/perpendicular bisector of B'AC'), we have $ \boxed{\angle B'PC' = 60^{\circ} + 2\alpha}$. Now
$ \angle BPC' = \angle BPB' + \angle B'PC' = \alpha + 60^{\circ} + 2\alpha = 60^{\circ} + 3\alpha$
$ \boxed{\boxed{\angle BPC' = 60^{\circ} + 3\alpha}}$
By the exterior angle theorem, $ \angle BQC' = \angle ABQ + \angle BAQ = \alpha + 60^{\circ}$, and since $ \triangle BQC'$ is isosceles, $ \boxed{\angle AC'B' = 60^{\circ} - \alpha/2}$. Also, since $ \angle C'B'B = 60^{\circ}$,
$ \angle PBC' =$
$ \angle PBB' - \angle C'BB' =$
$ \angle PBB' - (180^{\circ} - \angle CB'B - \angle BC'B') =$
$ \angle PBB' - (180^{\circ} - \angle CB'B - (\angle AC'B' - \angle AC'B')) =$
$ 180^{\circ} - 2\alpha - (180^{\circ} - 60^{\circ} - (60^{\circ} - (60^{\circ} - \alpha/2))) =$
$ 60^{\circ} - 3\alpha/2$
$ \boxed{\boxed{\angle PBC' = 60^{\circ} - 3\alpha/2}}$
The two double-boxed statements clearly imply that
$ \boxed{\boxed{\triangle BPC'\text{ is isosceles with }BP = PC'}}$ we can show that
$ \boxed{\triangle BPC'\text{ is isosceles with }BP = PC'}$.
Since QBC' is isosceles, the above boxed statement translates to
$ \boxed{QP\text{ bisects }\angle BQC}$
Now it is an easy exerciseBy the angle bisector theorem, $ BQ/QC = BP/PC$ and $ BA/AC = BP/PC$. So $ BQ/QC = BA/AC$, and combined with the law of sines,
$ \frac{\sin\angle QBC}{\sin\angle QCB} = \frac{\sin\angle ABC}{\sin\angle QCB} \implies \sin \angle QBC = \sin \angle ABC \implies \sin 2\alpha = \sin \alpha$
Therefore, $ 2\alpha + \alpha = 180^{\circ}$, so $ \angle ABC = 120^{\circ}$ and $ \angle ABC$ is obtuse. [In this special case where $ \angle BAC = 60^{\circ}$, the three vertices are collinear and $ ABC$ is degenerate]. to show that the above boxed statement, combined with the given, forces $ ABC$ to be obtuse (actually, degenerate). This is a contradiction.
So our initial assumption was false, and so B, P, C' are collinear. Now we can find the angles of ABC by noting that C = C' (both points are on lines BP and AQ - two lines intersect at one point (in Euclidean geometry )). This means that QB = QC, so that $ \alpha = \angle QCB$. Therefore, the three angles of ABC are $ 2\alpha$, $ \alpha$, and $ 60^{\circ}$. Solving for $ \alpha$ results in an $ \boxed{80-60-40\text{ triangle}}$.
math154
30.07.2010 02:24
vishalarul wrote: Now $ \angle BPC' = \angle BPB' + \angle B'PC' = \alpha + 60^{\circ} + 2\alpha = 60^{\circ} + 3\alpha$ The problem here is that $C'$ could be in between $Q$ and $C$, in which case we would have $360^\circ-\angle{BPC'}=\angle{BPB'}+\angle{B'PC'}$ instead. Of course, this can be fixed by directed angles (counterclockwise being positive). Basic angle chasing using the facts that $\triangle{AB'C'}$ is equilateral, $BP=BB'$, $QB=QC'$, and $PB'=PC'$ tells us that $\angle{PBC'}=\angle{BC'P}=60^\circ-(3\angle{ABQ}/2)$, so $PB=PC'$. But $PB'=PC'$, so $PB'=PC'=PB=BB'$. This implies that $\triangle{PBB'}$ is equilateral, so $\angle{ABQ}=\angle{BB'P}=60^\circ$. But this forces the sum of the angles in $\triangle{ABC}$ to be more than $180^\circ$, which is impossible. So we must have $\angle{APQ}=40^\circ$. Edit: I also have a trigonometric solution that is somewhat shorter than Virgil Nicula's. Letting $2x=\angle{ABQ}$, we find by the Law of Sines that (for convenience, take all angles to be in degrees) \[AB+\frac{AB\sin30}{\sin(150-4x)}=AB+BP=AQ+QB=\frac{AB\sin2x}{\sin(120-2x)}+\frac{AB\sin60}{\sin(120-2x)}.\]Dividing by $AB$ and simplifying, this is equivalent to \[1+\frac{1}{2\sin(4x+30)}=\frac{\sin2x+\sin60}{\sin(2x+60)}=\frac{2\sin(x+30)\cos(x-30)}{2\sin(x+30)\cos(x+30)}=\frac{\cos(x-30)}{\cos(x+30)}.\]Clearing denominators, this is equivalent to \[2\sin(4x+30)\cos(x+30)+\cos(x+30)=2\sin(4x+30)\cos(x-30),\]or \begin{align*} \cos(x+30) &= 2\sin(4x+30)[\cos(x-30)-\cos(x+30)]\\ &= 2\sin(4x+30)(2\sin{x}\sin30)\\ &= 2\sin(4x+30)\sin{x}\\ &= \cos(3x+30)-\cos(5x+30). \end{align*}Thus \[\cos(3x+30)=\cos(5x+30)+\cos(x+30)=2\cos(3x+30)\cos(2x).\]Note that $0<x<\angle{ABC}/4<45$. If $\cos(2x)=1/2$, then we must have $\angle{APQ}=2x=60$, making the sum of angles in $\triangle{ABC}$ greater than $180^\circ$. Thus $\cos(3x+30)=1\implies x=20$, so $\angle{ABC}=80$ and $\angle{ACB}=40$.
sayantanchakraborty
22.04.2014 21:04
Interestingly,both the geometry problems of IMO 2001 can be done by just a trigimetrical bash(though tiring and requires intelligence).Credit goes to those who will give a pure geometric approach.
utkarshgupta
03.12.2014 14:49
sayantanchakraborty wrote: Interestingly,both the geometry problems of IMO 2001 can be done by just a trigimetrical bash(though tiring and requires intelligence).Credit goes to those who will give a pure geometric approach. Actually I found it easier by pure geometry Solved it in less than half an hour The correct diagram is the key The trigonometric solutions would have taken me ages
mcdonalds106_7
16.05.2016 19:54
Let $\angle ABQ=\theta$. Note that $0< \theta < 60$. We will show that $\theta=40^\circ$ (Much of the calculation below is motivated by this answer).
Then, letting $AB=1$ gives that we must prove $1+\dfrac{\sin 30}{\sin(30+2\theta)}=\dfrac{\sin \theta}{\sin(60+\theta)}+\dfrac{\sin 60}{\sin(60+\theta)}$, or $\sin(60+\theta)(\sin 30 + \sin(30+2\theta))=\sin(30+2\theta)(\sin \theta +\sin 60)$. Now, we use the product to sum formula $\sin A \sin B = \dfrac{\cos(A-B)-\cos(A+B)}{2}$ to get, after some sign changes and grouping:
\[ [\cos(90-\theta)-\cos(30-2\theta)]+[\cos(-30)-\cos(150-3\theta)]+[\cos(30-\theta)-\cos(90-2\theta)]+[\cos(90-3\theta)-\cos 30]=0. \]
Now we use the sum to product formula $\cos A-\cos B=2\sin \left(\dfrac{A+B}{2}\right)\sin \left(\dfrac{A-B}{2}\right)$ to then get, after factoring:
\[ \sin\left(60-\dfrac{3\theta}{2}\right)\left[\sin\left(30+\dfrac{\theta}{2}\right)+\sin\left(-90+\dfrac{3\theta}{2}\right)+\sin\left(-30+\dfrac{\theta}{2}\right)+\sin\left(30-\dfrac{3\theta}{2}\right)\right]=0.\]
Clearly $\theta=40^\circ$ satisfies this, and if $\theta \neq 40^\circ$, then we must have $\left[\sin\left(30+\dfrac{\theta}{2}\right)+\sin\left(30-\dfrac{3\theta}{2}\right)\right]+\left[\sin\left(-90+\dfrac{3\theta}{2}\right)+\sin\left(-30+\dfrac{\theta}{2}\right)\right]=0$. By using the sum to product formula $\sin A+\sin B=2\sin \left(\dfrac{A+B}{2}\right)\cos \left(\dfrac{A-B}{2}\right)$, we then get
\[\sin\left(30-\dfrac{\theta}{2}\right)\cos \theta= \sin(60-\theta)\cos \left(30-\dfrac{\theta}{2}\right).\]
Using the $\sin$ and $\cos$ double angle-formulas allows us to simplify (Noting that $\theta<60^\circ$) to get $\cos \theta=1+\cos(60-\theta)$, but this never holds for $0<\theta<60$. Hence, $\theta=40^\circ$.
Durjoy1729
10.11.2018 18:15
It seems to me that it can be solved by construction.
illogical_21
01.05.2019 04:41
similar to #4
Let $X,Y$ be points on $AB,AC$ respectively such that $BP=PX$ and $BQ=QY.$ Then we have
$$AX=AB+BX=AB+BP=AQ+QB=AQ+QY=AY,$$so $AXY$ is equilateral.
Claim. $B,P,Y$ are collinear.
Proof. Suppose that $B,P,Y$ are not collinear. Note that $BPX,BQY$ are isoceles. Since $AP$ is an angle bisector of equilateral $AXY,$ we must have $PXY$ isoceles as well. Angle chasing, we have
\begin{align*}
\angle PYQ&=60^{\circ}-\angle XYP\\
&=60^{\circ}-\angle PXY\\
&=\angle PXB\\
&=\angle QBP,\\
\end{align*}then $XB=PB=PY=PX,$ which means $BXP$ is equilateral, or $\angle B=120^{\circ},$ a contradiction.
Then we have
\begin{align*}
180^{\circ}&=\angle BQY+\angle QYB+\angle YBQ\\
&=60^{\circ}+\frac32\angle B,\\
\end{align*}and $\boxed{\angle B=80^{\circ},\angle C=40^{\circ}},$ so we are done.
ubermensch
29.08.2019 16:40
Wow did I just do a G8 by a pure trig bash? Honestly, after applying the sine rule on both triangles and getting $\frac{\cos({m}+30)}{\cos({m}-30)}=\frac{\sin({30+4 m}) +\sin{30}}{\sin({30+4m})}$, it's almost mindless bashing... here is basically a sketch of my solution- First, letting $\angle{ABC}= \alpha$ and $m= \frac{\alpha}{4}$, just mark all angles and apply sine rule on triangles $APB$ and $AQB$ to get the aforementioned formula $\frac{\cos({m}+30)}{\cos({m}-30)}=\frac{\sin({30+4 m}) +\sin{30}}{\sin({30+4m})}$. Cross- multiply and simplify a bit to get a rather important step- $2 (\sin({60+3m})+ \sin{m}) \cos({2m}) = \sin{5m}+\sin({60+3m})$- this step is important because, if you weren't able to get a hint of what the angles may be through your diagram, the $60+3m$ is very telling of what the solutions could be... Juggling now with the algebra a bit, the next important step we get to is $\sin({60+3m}) \cdot (2 \cos({2m})-1) = \sin({5m})+\sin({m})-\sin({3m})=2 \sin({3m}) \cos({2m})-\sin({3m})=\sin({3m}) \cdot (2 \cos({2m})-1)$ Aha! Noticing that as $m< 30=>cos({2m}) \neq \frac{1}{2}$, we get $\sin({3m})=\sin({60+3m})=>6m=120=>m=20 => \alpha=80$ ! Hence our angles are $60,80,40$, and we're done.
I know it was a little ugly bash, but isn't this a little too easy for a G8, as there wasn't any real "Geometry" involved, at least in my solution...
AopsUser101
04.05.2020 03:02
Made with Blood, Sweat, and Tears Let $\angle ABQ = \angle BQC = x$. If $BP = \sin 30$, then by the Law of Sines, we know that $$\frac{BP}{\sin (30)} = \frac{AB}{\sin(150-2x)} \Longleftrightarrow AB = \sin (150 -2x)$$Furthermore, by the same reasoning as before, $$\frac{AB}{\sin (120-x)} = \frac{BQ}{\sin (60)} \Longleftrightarrow BQ = \frac{\sin (150 -2x) \sin (60)}{\sin (120-x)}$$$$\frac{AB}{\sin (120-x)} = \frac{AQ}{\sin (x)} \Longleftrightarrow AB= \frac{\sin (150 -2x) \sin (x)}{\sin (120-x)}$$It follows that since we want $BP + AB = AQ + QB$, we need: $$\frac{\sin (150-2x)(\sin(x) + \sin (60))}{\sin (120-x)} = \sin (30) + \sin(150 -2x)$$We know that $\sin (x) + \sin (60) = 2\sin \left(\frac{x + 60}{2} \right)\cos \left(\frac{x - 60}{2} \right)$ and $$\sin (120-x) = \sin (60 + x) = 2\sin \left(\frac{60 + x}{2} \right) \cos \left( \frac{60 + x}{2} \right)$$Therefore, we can reduce the problem to: $$\frac{\sin (150-2x)\left( 2\sin \left(\frac{x + 60}{2} \right)\cos \left(\frac{x - 60}{2} \right)\right)}{2\sin \left(\frac{60 + x}{2} \right) \cos \left( \frac{60 + x}{2} \right)} = \sin (30) + \sin(150 -2x) \Longleftrightarrow$$$$\frac{\sin (150 - 2x) \cos \left( \frac{x-60}{2} \right)}{\cos \left( \frac{x+60}{2} \right)} = \sin (30) + \sin(150 -2x) \Longleftrightarrow$$$$\frac{\cos \left( \frac{x-60}{2} \right)}{\cos \left( \frac{x+60}{2} \right)} = \frac{\sin (30)}{\sin(150-2x)} + 1$$This motivates us to consider: $$\frac{\cos \left( \frac{x-60}{2} \right)}{\cos \left( \frac{x+60}{2} \right)} - 1 = \frac{\cos \left( \frac{x-60}{2}\right) - \cos \left(\frac{x+60}{2} \right)}{\cos \left(\frac{x+60}{2} \right)} = \frac{-2 \sin \left(\frac{x}{2} \right) \sin (-30)}{\cos \left(\frac{x+60}{2} \right)}$$Putting this into our equation, we see that: $$\frac{-2 \sin \left(\frac{x}{2} \right) \sin (-30)}{\cos \left(\frac{x+60}{2} \right)} = \frac{\sin (30)}{\sin(150-2x)} \Longleftrightarrow$$$$\frac{2 \sin \left(\frac{x}{2} \right)}{\cos \left(\frac{x+60}{2} \right)} = \frac{1}{\sin (150-2x)} \Longleftrightarrow$$$$2\sin \left(\frac{x}{2} \right) \sin (150 -2x) = \cos \left(\frac{x}{2} + 30 \right)$$Multiplying both sides by $2\cos \left(\frac{x}{2} \right)$, we get: $$2\sin (x) \sin (150 -2x) = 2\cos \left(\frac{x}{2} + 30 \right) \cos \left(\frac{x}{2} \right) \Longleftrightarrow $$$$- \cos \left(150 - x \right) - \cos (150 -3x) = \cos(x+30) + \cos(30) \Longleftrightarrow$$$$\cos(150 - 3x) = \cos (30) \Longleftrightarrow x = 40$$This means that the only such triangle has angles $60,40,80$.
GeronimoStilton
12.06.2020 05:04
Let $\angle ABQ=x$, so $\angle ABC = 2x$. By the angle bisector theorem and the Law of Sines, \[AB+\frac{AB}{AB+AC}\cdot BC = AB+BP=AQ+BQ = AB \cdot \frac{\sin \angle ABQ+\sin \angle BAQ}{\sin \angle AQB}.\]Divide through by $AB$ and apply the Law of Sines again to get \[\frac{\sin 30^\circ+\sin (2x+30^\circ)}{\sin (2x+30^\circ)} = 1+\frac{\sin 60^\circ}{2\sin (2x+30^\circ)\cos (30^\circ)} = \]\[1+\frac{\sin 60^\circ}{\sin (2x+60^\circ)+\sin 2x} = \frac{\sin (60^\circ+2x)+\sin 2x+\sin 60^\circ}{\sin (60^\circ+2x)+\sin 2x}=\frac{\sin \angle ACB+\sin \angle ABC+\sin \angle BAC}{\sin \angle ACB+\sin \angle ABC}=\]\[\frac{AB+AC+BC}{AB+AC}=\frac{\sin \angle ABQ+\sin \angle BAQ}{\sin \angle AQB} = \frac{\sin x+\sin 60^\circ}{\sin (60^\circ+x)} = \frac{2\sin (x/2+30^\circ)\cos (x/2-30^\circ)}{2\sin (x/2+30^\circ)\cos (x/2+30^\circ)}=\]\[\frac{\cos (x/2-30^\circ)}{\cos (x/2+30^\circ)}.\]Cross-multiply to obtain \[\sin (30^\circ)\cos (x/2+30^\circ)+\sin (2x+30^\circ)\cos (x/2+30^\circ) = \sin (2x+30^\circ)\cos (x/2-30^\circ).\]Rewrite this as \[\sin (30^\circ)\cos (x/2+30^\circ)=\sin (2x+30^\circ)(\cos (x/2-30^\circ) + \cos (150^\circ - x/2)) = \]\[2\sin (2x+30^\circ)\cos (60^\circ)\cos (x/2-90^\circ) = 2\sin (2x+30^\circ)\cos (60^\circ)\sin (x/2).\]Divide through by $\sin (30^\circ)=\cos (60^\circ)$ to get \[\cos (x/2+30^\circ) = 2\sin (2x+30^\circ)\sin (x/2) = \cos (3x/2+30^\circ) - \cos (5x/2+30^\circ).\]Rearrange as \[\cos (3x/2+30^\circ) = \cos (x/2+30^\circ)+\cos (5x/2+30^\circ) = 2\cos (3x/2+30^\circ)\cos (x).\]This implies that either $\cos x = 1/2$ or $\cos (3x/2+30^\circ) = 0$. The first implies $x=60^\circ \implies \angle ABC = 120^\circ \implies \angle ACB = 0^\circ$, absurd. Thus, we have that $3x/2+30^\circ \equiv 90^\circ\pmod{180^\circ}$, so $x/2\equiv 20^\circ\pmod{60^\circ} \implies \angle ABC \equiv 80^\circ \pmod{240^\circ}$. Thus, we get $\angle ABC = 80^\circ$ and $\angle ACB = 40^\circ$.
jelena_ivanchic
24.05.2021 12:46
Let $\angle ABQ=x$, so $\angle ABC = 2x$.Also let $QB=\sin 60.$ ( Scaling won't affect the angles)
Here we have, $AB+BP=AQ+QB.$ We will Note that by angle bisector theorem we have $AQ=\frac{AB}{AB+AC}\cdot BC.$
Now we find all the lenghts in terms of $\sin.$
Using $\sin$ law in $QBA,$ we get,
$$\frac{QB}{\sin 60}=\frac{AB}{\sin (120-x)}=\frac{AQ}{\sin x}.$$Since we had $QB=\sin 60\implies AB=sin(120-x),~~AQ=sin(x).$.
Now, we find $AC,BC.$ Again using $\sin$ law on $\Delta ABC,$
we get $$\frac{BC}{\sin 60}=\frac{AB}{\sin(120-x)}=\frac{\sin(120-x)}{\sin(120-x)}\implies BC=\frac{\sin(120-x)\cdot \sin 60 }{\sin(120-2x)}$$Similarly, we get $$ \frac{AC}{sin (2x)}=\frac{AB}{\sin(120-2x)}\implies AC=\frac{\sin(120-x)\cdot sin(2x)}{\sin(120-2x)}.$$
Now putting all the values we get
$$sin\left(60\right)+sin\left(x\right)=sin\left(120-x\right)\left(1+\frac{\left(\frac{sin\left(120-x\right)\left(sin\:60\right)}{sin\left(120-2x\right)}\right)}{\left(sin\left(120-x\right)+\left(\frac{sin\left(120-x\right)sin\left(2x\right)}{sin\left(120-2x\right)}\right)\right)}\right) $$Simplifying the LHS, we get
$$sin\left(120-x\right)\left(1+\frac{\left(\frac{sin\left(120-x\right)\left(sin\:60\right)}{sin\left(120-2x\right)}\right)}{\left(sin\left(120-x\right)+\left(\frac{sin\left(120-x\right)sin\left(2x\right)}{sin\left(120-2x\right)}\right)\right)}\right)$$$$=\sin(120-x)+\frac{\left(sin\left(120-x\right)\left(sin\:60\right)\right)}{\left(sin\left(120-2x\right)+sin\left(2x\right)\right)}$$$$=sin\left(60+x\right)\left(1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}\right)$$Note that $\sin(60+x)=2\sin(\frac{60+x}{2})\cos(\frac{60+x}{2}).$
Also simplifying the RHS, we get $sin\left(60\right)+sin\left(x\right)=2\sin(\frac{x+60}{2})\cos(\frac{x-60}{2}).$
Since LHS=RHS we cancel out $\sin(\frac{x+60}{2})\ne 0.$
We get $$cos\left(\frac{x}{2}-30\right)=cos\:\left(\frac{x}{2\:}+30\right)\left(1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}\right) $$So we have
$$ \frac{cos\left(\frac{x}{2}-30\right)}{cos\:\left(\frac{x}{2\:}+30\right)}=1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}$$Note that here RHS is
$$ 1+\frac{\left(sin\left(60\right)\right)}{\left(sin\left(60+2x\right)+sin\left(2x\right)\right)}$$$$=1+\frac{sin\left(60\right)}{\left(2sin\left(2x+30\right)+cos\left(30\right)\right)}$$$$1+\frac{1}{2sin\left(2x+30\right)} $$So we have $$ \frac{cos\left(\frac{x}{2}-30\right)}{cos\:\left(\frac{x}{2\:}+30\right)}=\frac{1+sin\left(2x+30\right)}{2sin\left(2x+30\right)}$$Cross multiplying, we get
$$cos\left(\frac{x}{2}-30\right)2sin\left(2x+30\right)=cos\:\left(\frac{x}{2\:}+30\right)+2sin\left(2x+30\right)cos\:\left(\frac{x}{2\:}+30\right) $$$$ =sin\left(\frac{5x}{2}\right)+sin\left(60+\frac{3x}{2}\right)=cos\:\left(\frac{x}{2\:}+30\right)+sin\left(\frac{5x}{2}+60\right)+sin\left(\frac{3x}{2}\right)$$Now taking $x/2=y,$ we get
$\sin(5y+60)+\sin(60-y)+\sin(3y)=\sin(60+3y)+\sin(5y)\implies \sin(5y+60)-\sin(5y)+\sin(60-y)=\sin(60+3y)-\sin(3y).$
So, we have,
$$\cos (5y+30)+\sin(60-y)=\cos(3y+30)\implies \sin(60-5y)+\sin(60-y)=\cos(3y+30)$$$$\implies 2\cdot \cos(3y+30)\cos(60-2y)=\cos (3y+30). $$
Note that here $0<y\le 30$
So $\cos(6-2y)=1/2$ is eliminated. Hence $\cos(30+3y)=0\implies y=20=x/2\implies \angle ABC=2x=80\implies \angle ACB=40.$
Maths_1729
24.05.2021 15:58
orl wrote: Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle? Nice problem. Here's another solution. Take Point $D$ such that $\angle ABD=60^\circ\implies \triangle ABD$ is equilateral, Let $AP\cap BD=H$ So $PD=BP$ Take Point $E\in BC$ such that $BE>BD$ and $DE=PD=BP$ Also take Point $F$ such that $BF>BA$ and $AF=BP$ and observe we have another equilateral $\triangle BFE$ Now take point $G\in BQ$ Such that $BG>BQ$ and $QG=AQ$ Claim 1-: $B$ is circumcenter of $\triangle FGE$ Prove-: it is Trivial observation As $BG=BE=BF$ Take Point $J$ Such that $BG=BJ$ Claim 2-: $GJ=FG=JE$ Prove-: First observe $\triangle BFG\cong \triangle BJE \implies FG=JE, \angle GFE=\angle JEF$ and also $FE||GJ$ So $GJEF$ is cyclic isosceles Trapezium. So $GJ=FG=JE$ Claim 3-: $\angle FBG=20^\circ$ Prove-: By Claim $2$ We get $\angle FEG=\frac{\angle FEJ}{2}$ and By Claim $1$ we get $\angle GEF=\frac{\angle FBG}{2}$ and $\angle EFG=\angle FEJ=\frac{60-\angle FBG}{2}$ So we get $\frac{\angle FBG}{2}=\frac{60-\angle FBG}{4}\implies \angle FBG=20^\circ$ Now By claim $3$ we get angles of $\triangle ABC$ are $\{40^\circ, 60^\circ, 80^\circ\}$ $\blacksquare$ Edit-: 200th Post in HSO
Marinchoo
02.02.2022 03:25
I think bashing on this one is an insult to the problem. Here is a synthetic solution: Let $B_{1}\in AC$ and $C_{1}\in AB$, so that $QB_{1}=QB$ and $BP=BC_{1}$ and $Q$ is between $A$ and $B_{1}$ and $B$ is between $A$ and $C_{1}$, and we have that: $$AB_{1}=AQ+QB_{1}=AQ+QB=AB+BP=AB+BC_{1}=AC_{1}\Longrightarrow \angle C_{1}AB_{1}=60^{\circ}, AB_{1}=AC_{1}\Longrightarrow \triangle AB_{1}C_{1}\text{ is equilateral}$$Notice that $\angle BC_{1}P=\angle BPC_{1}=\frac{\beta}{2}$ and $P$ lies on the angle bisector in an equilateral triangle, so $\triangle PAC_{1}\cong \triangle PAB_{1}$ $\Longrightarrow \angle PB_{1}A=\angle PC_{1}A=\frac{\beta}{2}$. Now assume that $P\not\in BB_{1}$. Then $QB=QB_{1}$ by construction and $\angle PBQ=\angle PB_{1}Q$, so since $P$ lies in the interior of $\triangle QBB_{1}$, then that'd mean that $P$ lies on the perpendicular bisector of $BB_{1}$, so $PQ\perp BB_{1}\Longrightarrow \angle BQP=\angle B_{1}QP$, so $P$ is the $A$-excenter for $\triangle ABQ$, which implies that $\angle PBQ=\angle PBC_{1}\Longrightarrow \frac{3}{2}\beta=180^{\circ}\Longrightarrow \beta=120^{\circ}$, which implies that $\gamma=0$, which is obviously impossible. Therefore $P\in BB_{1}\Longrightarrow C\equiv B_{1}\Longrightarrow \gamma=\frac{\beta}{2}$, so $\beta=80^{\circ}$ and $\gamma=40^{\circ}$ and we're done!
Basically whenever we're faced with equality of (sum of) segments like in this problem we try to translate the segments onto one or two lines and then angle chase, as shown above. Quite a beautiful problem in my opinion.
asdf334
22.02.2022 18:34
lol. Let $\cos \theta=x$ where $\theta=\angle ABQ=\angle CBQ$. Then we must solve $$3(4x^3-2x^2-2x+1)^2=(1-x^2)(4x^2-6x+2)^2$$and we can factor out $(2x-1)^3$ since $x=\frac{1}{2}$ is extraneous which gives $$8x^3-6x+1=0\iff \cos 3\theta=-\frac{1}{2}$$and since $\theta<60^{\circ}$ this means that $3\theta = 120^{\circ}$ and $\theta = 40^{\circ}$. Answer is $\angle A, \angle B, \angle C = 60^{\circ}, 80^{\circ}, 40^{\circ}$.
asdf334
22.02.2022 18:44
the sextic comes from bashing out $$1+\frac{\sin 30^{\circ}}{\sin(2\theta+30^{\circ})}=\frac{\sin \theta + \sin 60^{\circ}}{\sin(\theta + 60^{\circ})}$$
awesomeming327.
05.03.2022 03:30
Yolo. Let $D$ be on $AB$ extended such that $ACD$ is equilateral. Let $C'$ be on $AC$ such that $QB=QC'$ and $A,C'$ are on opposite sides of $Q.$ Let $D$ be on $AB$ such that $BP=BD$ and $A,D$ are on opposite sides of $B.$ Note that $AB+BP=AQ+QB$ implies $AC'=AB+BD=AQ+QC'=AD$ so $AC'D$ is equilateral. Since $AP$ is an angle bisector, $\angle AC'P=\angle ADP=\angle BPD=\frac{1}{2}\angle PBA=\angle PBQ.$ However, we have $QC'=QB,QP=QP,$ and $\angle QC'P=\angle QBP,$ so by quick sine law on $\triangle QPC',\triangle QPB$, we get that $\sin QPC'=\sin QPB.$ If they are supplementary then $C=C'$ and so $\angle BCD=\angle PDC=\frac{1}{2}\angle BPD=\frac{1}{2}\angle ACB$ and $\angle BCD+\angle ACB=60^\circ$ so $\angle ACB=40^\circ,\angle ABC=80^\circ.$ If they are the same then $PB=PC=PD,$ so $P$ is the circumcenter of $BCD$ which is impossible because $\angle CDB\neq 90^\circ.$ Thus, our only solution is $\{40^\circ,60^\circ,80^\circ\}$
Mogmog8
11.06.2022 03:34
Let $X$ and $Y$ be the points on ray $\overrightarrow{AB}$ and segment $\overline{AC}$ such that $BX=BP$ and $BQ=QY,$ respectively. Then, $\triangle APX\cong\triangle APY$ so $$\alpha=\angle AYP=\angle PXA=90-\tfrac{1}{2}\angle XBP=\angle PBQ=\angle QBP.$$Suppose FTSOC that $Y$ does not lie on $\overline{PB}.$ Then, $$\angle PBY=\angle PBQ-\angle YBQ=\angle QYP-\angle QYB=\angle QYP$$so $PX=PY=PB$ and $\alpha=60.$ Then, $\angle A+\angle B=60+2\alpha=180,$ meaning $\angle C=0,$ a contradiction. Hence, $B,C,$ and $P$ are collinear meaning $Y=C.$ Therefore, the sum of the angles of $\triangle ABC$ is $60+3\alpha$ so $\alpha=40$ and $\angle A=60,\angle B=80,\angle C=40.$ $\square$
asdf334
04.08.2022 03:58
pbed fuzi L wrote: aime p14 lvl bash in imo p5??? WOLOG let $R = \frac 12$ and $\angle ABQ = \angle QBC = x$. By Law of Sines on $\triangle ABP$, $BP = \frac{\sin C}{2\sin(150^{\circ}-2x)}$, and by LoS on $\triangle ABQ$, $AQ = \frac{\sin C\sin x}{\sin(120^{\circ}-x)}, QB = \frac{\frac{\sqrt{3}}{2}\sin C}{\sin(120^{\circ}-x}$. Thus, we can encode the given as\begin{align*}AB + BP &= AQ + QB \\ c + \frac{\sin C}{2\sin(150^{\circ}-2x)} &= \frac{\sin C\sin x}{\sin(120^{\circ}-x)} + \frac{\frac{\sqrt{3}}{2}\sin C}{\sin(120^{\circ}-x} \\ 1 + \frac{1}{2\sin(150^{\circ} - 2x)} &= \frac{\sin x + \sin 60^{\circ}}{\sin(120^{\circ}-x)}\\ 1 + \frac{1}{2\sin(150^{\circ} - 2x)} &= \frac{\cos\left(\frac x2 - 30^{\circ}\right)}{\cos\left(\frac x2 + 30^{\circ}\right)} \\ 2\cos\left(\frac x2 - 30^{\circ}\right)\sin(150^{\circ}-2x)&= 2\cos\left(\frac x2 + 30^{\circ}\right)\sin(150^{\circ}-2x) + \cos\left(\frac x2 +30^{\circ}\right) \\ \sin\left(120^{\circ}-\frac{3x}{2}\right) + \sin \left(\frac{5x}{2}\right) &= \sin\left(\frac{3x}{2}\right) + \sin\left(120^{\circ} - \frac{5x}{2}\right) + \cos\left(\frac x2 + 30^{\circ}\right) \\ \sin \left(\frac{5x}{2}\right) -\sin\left(\frac{3x}{2}\right) &= \sin\left(120^{\circ} - \frac{5x}{2}\right) - \sin\left(120^{\circ}-\frac{3x}{2}\right) + \cos\left(\frac x2 + 30^{\circ}\right) \\ 2\cos(2x)\sin\left(\frac x2\right) &= -2\cos(120^{\circ}-2x)\sin\left(\frac x2\right) + \cos\left(\frac x2 + 30^{\circ}\right) \\ 2\sin\left(\frac x2\right)\left[\cos(2x) + \cos(120^{\circ} - 2x)\right] &= \cos\left(\frac x2 + 30^{\circ}\right) \\ 2\sin\left(\frac x2\right)\cos(2x-60^{\circ}) &= \cos\left(\frac x2 + 30^{\circ}\right) \\ \sin x \cos(2x-60^{\circ}) &= \cos\left(\frac x2 + 30^{\circ}\right) \cos\left(\frac x2\right) \\ \sin(3x-60^{\circ}) &= \cos(30^{\circ}) \\ x &= 40^{\circ}, 60^{\circ}, 80^{\circ}.\end{align*}And we're FINALLY done. godermit this problem sucks
YaoAOPS
16.08.2022 22:49
Let $x = \angle ABQ$ Note that by law of sines, $$\frac{AQ}{\sin x} = \frac{BQ}{\sin 60} = \frac{c}{\sin\angle AQB}$$and $$\frac{BP}{\sin 30} = \frac{c}{\sin\angle APB}$$
As such, \begin{align*}AQ + QB = AB + BP &\iff \\ \frac{c}{\sin\angle AQB}(\sin x + \sin 60) = \frac{c}{\sin\angle APB}(\sin\angle 30 + \sin 2x + 30) &\iff \\ \frac{\sin x + \sin 60}{\sin(x + 60)} = 1 + \frac{1}{2\sin 2x + 30} \end{align*}
Then, since $\sin x + \sin 60 = \sin (\frac{x}{2} + 30) \cos(\frac{x}{2} - 30)$, $$\frac{\sin x + \sin 60}{\sin(x + 60)} = \frac{\sin (\frac{x}{2} + 30) \cos(\frac{x}{2} - 30)}{2\sin(\frac{x}{2} + 30)\cos(\frac{x}{2} + 30)} = \frac{\cos(\frac{x}{2} - 30)}{\cos(\frac{x}{2} + 30)} $$
Then, the equation becomes \begin{align*}
\frac{\cos(\frac{x}{2} - 30) - \cos(\frac{x}{2} + 30)}{\cos(\frac{x}{2} + 30)} = \frac{1}{2\sin(2x + 30)} &\iff \\ \frac{\sin \frac{x}{2}}{\cos(\frac{x}{2} + 30)} = \frac{1}{2\sin(2x + 30)} &\iff \\
2\sin(2x + 30)\sin \frac{x}{2} = \cos(\frac{x}{2} + 30) &\iff \\
\cos\left(\frac{3x}{2} + 30\right) = \cos\left(\frac{5x}{2} + 30\right) + \cos\left(\frac{x}{2} + 30\right)
\end{align*}
Substitute $a = \frac{3x}{2} + 30$ to get $\cos a = 2 \cos a \cos x$. Then either $x = 60$ or $a = 90$. The first case implies $\angle C = 0$, impossible, so $\frac{3x}{2} + 30 = 90$ and $x = 40$. Thus, $\angle A = 60$, $\angle B = 80$, $\angle C = 40$.
huashiliao2020
05.06.2023 06:19
orl wrote: Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle? Wow!! A G8!! Can someone check if this is right (especially the second solution)? WLOG BP=1.
Exactly, for this solution I am showing no work. I did indeed use wolfram alpha
Simplifying this, we get angles of 40, 80, and 60 degrees. Second solution: (Latexing this was a NIGHTMARE.) In these types of problems with little points and information, we must surely make a construction (for once I will make one!). The easiest way to do so is to make two segments' sum into two segments' sum which are collinear, since it is much simpler that way. Then we let B' on AB s.t. $BP=BB'$, C' on AC s.t. $QB=QC'$ (Remark: if this wasn't geogebra it would be hard to guess what type of diagram meaning it would be hard to tell C coincides with C'). We have $AB+BB'=AB+BP=AQ+QB=AQ+QC'$, or $AB'=AC'$ so $\triangle AB'C'$ is equilateral. Now, suppose B, P, and C' are not collinear. Then, we have $\angle ABC=2b$ gives $\angle BB'P=\angle BPB'=b$. By corresponding angles, BQ//B'P, and $\angle PB'C'=60-b$, and because B'PC' is isosceles, we have $\boxed{\angle{B'PC'}=60+2b}$. $\boxed{\angle BPC'}=\angle BPB'+\angle B'PC'=b+60+2b=\boxed{60+3b}$, $\angle BQC'=60+b$, and since BQC' is isosceles, $\angle BC'A=\angle BC'Q=\boxed{60-b/2}$; Finally, $\boxed{\angle{PBC'}}$=<PBB'-<C'BB'=<PBB'-(180-<BC'B'-<BB'C')=<PBB'-(180-<BB'C'-(<AC'B'-<BC'A))=$180-2b-(180-60-(60-(60-b/2)))=\boxed{60-3b/2}$. Then $\boxed{\angle BC'P=180-(60-3b/2)-(60+3b)=60-3b/2}$, so $\boxed{BP=PC'}$. We knew that QBC' was isosceles, so QP bisects <BQC, which gives BA/AC=BP/PC=BQ/QC. By Law of Sines, we have sin<QBC/sin<QCB=QC/QB=AC/AB=sin<ABC/sin<ACB, which gives sinb=sin<QBC=sin<ABC=sin2b, or 2b+b=180 b=60, and <ABC=120. However, this makes a degenerate triangle ABC, which is a contradiction. Then B, P, C' are collinear, so C=C' and therefore BQ=QC'=QC which means <QCB=<QBC=b, and 2b+b=180-60=120, so b=40 and then our angles are 40, 60, and 80 degrees. In fact, while the synthetic way is harder to see where to go, whereas trig you always know what to do, the synthetic would probably be quicker even though the solutions length is longer, because it is just a bunch of simple angle chasing, with one obvious construction. $\blacksquare$
sixoneeight
06.11.2023 06:07
Solved with popop614 Let $P'$ be the reflection of $P$ over the exterior angle bisector of $B$ in triangle $ABC$, and let $B'$ be the reflection of $B$ over the angle bisector of $\angle BQC$. Then, the length condition means that $AB'P'$ is equilateral. Assume that $BPB'$ are not collinear. Now, note that $P$ is symmetric with respect to $AP'$ and $AB'$, so $\triangle PP'B'$ is isosceles. It is also symmetric with respect to $QB$ and $QB'$, so $BPB'$ is isosceles, implying $BP'=BP=PB'=PP'$. In other words, $BPP'$ is equilateral, so $\angle ABC = 120$, implying $ABC$ is degenerate. Therefore, $BPB'$ collinear, so $B'=C$. Therefore, $2\angle BCA = \angle ABC$. This implies that $ABC$ has angles $40$, $60$, and $80$.
InterLoop
21.04.2024 18:12
one of my first synthetic sols to 'those' geos
Construct points $P'$ and $Q'$ on $AB$ and $AC$ such that $AP' = AB + BP' = AB + BP$, and $AQ' = AQ + QQ' = AQ + QB$. Note that thus $AP'Q'$ is equilateral. Also now note that $BPQ$, $BQY$, and $PXY$ are isosceles. We obtain $\angle PQ'Q = 60^{\circ} - \angle PQ'P' = 60^{\circ} - \angle PP'Q' = \angle PP'B = \angle PBQ$, which gives a contradiction (triangle is line) unless $Q'$ lies on $BP$, thus $Q' = C$. Thus we obtain $60^{\circ} + 3\angle C = 180^{\circ}$, implying $\angle C = 40^{\circ}$ and $\angle B = 80^{\circ}$.
joshualiu315
09.05.2024 12:01
imagine knowing synthetic oops Set $\angle ABQ = x$, and WLOG let $AB=1$. Using Law of Sines, we see that \[BP = \frac{\sin 30^\circ}{\sin (150^\circ-2x)},\]\[AQ = \frac{\sin x}{\sin (120^\circ-x)},\]\[QB = \frac{\sin 60^\circ}{\sin (120^\circ-x)}.\] Thus, the given condition translates to \[1+\frac{\sin 30^\circ}{\sin (150^\circ-2x)} = \frac{\sin x}{\sin (120^\circ-x)} + \frac{\sin 60^\circ}{\sin (120^\circ-x)},\] whence the RHS simplifies to \[\frac{\sin x + \sin 60^\circ}{\sin (60^\circ+x)} = \frac{2 \sin (\tfrac{1}{2} (x+60^\circ)) \cos (\tfrac{1}{2} (x-60^\circ))}{2 \sin (\tfrac{1}{2} (x+60^\circ)) \cos (\tfrac{1}{2} (x+60^\circ))}.\] Now, suppose that $y = \tfrac{x}{2}$ for brevity; we now have \begin{align*} &\phantom{\iff} 1+\frac{\sin 30^\circ}{\sin (150^\circ-4y)} = \frac{\cos (y-30^\circ)}{\cos (y+30^\circ)} \\ &\iff \frac{\sin 30^\circ}{\sin (150^\circ-4y)} = \frac{\cos (y-30^\circ)}{\cos (y+30^\circ)} -1 \\ &\iff \frac{\sin 30^\circ}{\sin (150^\circ-4y)} = \frac{\cos (y-30^\circ) - \cos (y+30^\circ)}{\cos (y+30^\circ)} \\ &\iff \frac{\sin 30^\circ}{\sin (150^\circ-4y)} = \frac{\sin y}{\cos (y+30^\circ)}. \end{align*} This is equivalent to \begin{align*} \cos (y+30^\circ) &= 2 \sin y \sin (150^\circ - 4y) \\ &= \cos (5y-150^\circ) - \cos (150^\circ - 3y) \\ &= - \cos (5y+30^\circ) + \cos (3y + 30^\circ). \end{align*} Rearranging, we get \[\cos (3y+30^\circ) = \cos (y+30^\circ) + \cos (5y+30^\circ) = 2 \cos (3y+30^\circ) \cos (2y)\]\[\implies \cos (3y+30^\circ) (2 \cos (2y) - 1) = 0.\] The latter case does not produce any solutions as we require $\angle B < 120^\circ$ and the former case gives $y = 20^\circ$, upon which we find $\angle B = 80^\circ$ and $\angle C = 40^\circ$ as the only solutions.
ehuseyinyigit
03.08.2024 04:23
The problem is equilavent to this Trigonometric Equation for $\alpha \in (0,60^{\circ})$.
Markas
23.09.2024 01:02
Let AB = c. Now by law of sines on $\triangle ABP$, we get that $BP = \frac{c \cdot \sin 30}{\sin (150 - 2\beta)}$. Also by law of sines on $\triangle ABQ$, we get that $AQ = \frac{c \cdot \sin \beta}{\sin (120 - \beta)}$ and $BQ = \frac{c \cdot \sin 60}{\sin (120 - \beta)}$ $\Rightarrow$ since AB + BP = AQ + QB, we have that $c + \frac{c \cdot \sin 30}{\sin (150 - 2\beta)} = \frac{c \cdot \sin \beta}{\sin (120 - \beta)} + \frac{c \cdot \sin 60}{\sin (120 - \beta)}$ $\Longleftrightarrow$ $1 + \frac{\sin 30}{\sin (150 - 2\beta)} = \frac{\sin \beta + \sin 60}{\sin (120 - \beta)}$ $\Longleftrightarrow$ $\frac{\sin (150 - 2\beta) + \sin 30}{\sin (150 - 2\beta)} = \frac{\sin \beta + \sin 60}{\sin (120 - \beta)}$ $\Longleftrightarrow$ $\frac{\sin (150 - 2\beta) \cdot (\sin \beta + \sin 60)}{\sin (120 - \beta)} = \sin 30 + \sin (150 - 2\beta)$. Since $\sin (x) + \sin 60 = 2\sin (\frac{\beta + 60}{2}) \cdot \cos (\frac{\beta - 60}{2})$ and $\sin (120 - \beta) = \sin (60 + \beta) = 2\sin(\frac{60 + \beta}{2}) \cdot \cos (\frac{60 + \beta}{2})$, we can reduce the problem to $\frac{\sin (150 - 2\beta) \cdot ( 2\sin (\frac{\beta + 60}{2}) \cdot \cos (\frac{\beta - 60}{2}))}{2\sin (\frac{60 + \beta}{2}) \cdot \cos (\frac{60 + \beta}{2})} = \sin 30 + \sin(150 - 2\beta)$ $\Longleftrightarrow$ $\frac{\sin (150 - 2x) \cdot \cos (\frac{\beta - 60}{2})}{\cos (\frac{\beta + 60}{2})} = \sin 30 + \sin(150 - 2\beta)$ $\Longleftrightarrow$ $\frac{\cos (\frac{\beta - 60}{2})}{\cos (\frac{\beta + 60}{2})} = \frac{\sin 30}{\sin(150 - 2\beta)} + 1$. Now we get $\frac{\cos (\frac{\beta - 60}{2})}{\cos (\frac{\beta + 60}{2})} - 1 = \frac{\cos (\frac{\beta - 60}{2}) - \cos (\frac{\beta + 60}{2})}{\cos (\frac{\beta + 60}{2})} = \frac{-2 \sin (\frac{\beta}{2}) \cdot \sin (-30)}{\cos (\frac{\beta + 60}{2})}$ $\Rightarrow$ we get $\frac{-2 \sin (\frac{\beta}{2}) \cdot \sin (-30)}{\cos (\frac{\beta + 60}{2})} = \frac{\sin 30}{\sin(150 - 2\beta)}$ $\Longleftrightarrow$ $\frac{2\sin (\frac{\beta}{2})}{\cos (\frac{\beta + 60}{2})} = \frac{1}{\sin (150 - 2\beta)}$ $\Longleftrightarrow$ $2\sin (\frac{\beta}{2}) \cdot \sin (150 - 2\beta) = \cos (\frac{\beta}{2} + 30)$. Multiplying by $2\cos (\frac{\beta}{2})$, we get $2\sin (\beta) \cdot \sin (150 - 2\beta) = 2\cos (\frac{\beta}{2} + 30) \cdot \cos (\frac{\beta}{2})$ $\Longleftrightarrow$ $- \cos (150 - \beta) - \cos (150 - 3\beta) = \cos(\beta + 30) + \cos 30$ $\Longleftrightarrow$ $\cos(150 - 3\beta) = \cos 30$ $\Longleftrightarrow$ $\beta = 40^{\circ}$ $\Rightarrow$ $\angle ABC = 2\beta = 80^{\circ}$ and $\angle ACB = 120 - 2\beta = 40^{\circ}$ $\Rightarrow$ $\angle A = 60^{\circ}$, $\angle B = 80^{\circ}$, $\angle C = 40^{\circ}$ and we are ready.
