USA IMO

First Solution. (by Reid Barton and Gabriel Carroll) Extend segment $AB$ through $B$ to $R$ so that $BR=BP$, and construct $S$ on ray $AQ$ so that $AS=AR$.



Lemma 1. Points $B$, $P$, $S$ are collinear, and consequently, $S$ coincides with $C$.

Proof. Because $BR=BP$, triangle $BPR$ is isosceles with base angles $$\angle BRP=\angle RPB=(180^{\circ }-\angle PBR)/2=x=\angle QBP.(2)$$ Note that $AS=AR$ and $\angle RAS=\angle BAC=60^{\circ }$, implying that triangle $ARS$ is equilateral. Since line $AP$ bisects $\angle RAS$, $R$ and $S$ are symmetric with respect to line $AP$. Thus, $$PR=PS(3)$$ and $\angle ARP=\angle PSA$, or $\angle BRP=\angle PSQ$. By (2), we have $$\angle QBP=\angle BRP=\angle PSQ.(4)$$

Because $AQ+QS=AB+BR=AB+BP=AQ+QB$, $QS=QB$. Hence, triangle $BQS$ is isosceles with $$\angle BSQ=\angle QBS.(5)$$

Now, assume to the contrary that triangle $BPS$ is nondegenerate. Then either $AC<AS$, as in the first diagram below, or $AC>AS$, as in the second diagram.



In either diagram, combining (4) and (5) gives $$\angle PBS=|\angle QBP-\angle QBS|=|\angle PSQ-\angle BSQ|=\angle PSB,$$ that is, triangle $PBS$ is isosceles with $PB=PS$. By (3), it follows that $PB=PS=PR$. Hence, triangle $BPR$ is equilateral. But then $\angle ABC=180^{\circ }-\angle CBR=120^{\circ }$, and by (1), $y=0^{\circ }$, which is absurd. Therefore, our assumption was wrong and $B,P,S$ are collinear. Consequently, $S=C$. ■

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Second Solution. (By Zhiqiang Zhang, China) Construct $R$ on ray $AB$ beyond $B$ such that $BR=BP$, and construct $S$ on ray $AQ$ beyond $Q$ such that $QS=QB$. Let lines $BS$ and $AP$ intersect at $D$. Note that $AR=AB+BP=AQ+QB=AS$. Then both triangles $BPR$ and $BQS$ are isosceles with $$\angle BRP=\angle RPB=\frac{\angle ABC}{2}=x=\angle QBP(6)$$ and $$\angle QBS=\angle BSQ=\frac{\angle BQA}{2}=\frac{180^{\circ }-60^{\circ }-x}{2}=60^{\circ }-\frac{x}{2}.(7)$$

Because $y>0^{\circ }$, equation (1) implies $x<60^{\circ }$. Because $AR=AS$ and $\angle RAS=60^{\circ }$, we also have that $60^{\circ }=\angle ARS=\angle BRS$. Using these relations as well as (6), we find that $\angle BRP=x<60^{\circ }=\angle BRS$, implying that $P$ is inside triangle $ARS$. We now consider two cases.

i. $C=S$. Since $BQ=QC$, $y=\angle BCQ=\angle QBC=x$, so by (1), $\angle ABC=80^{\circ }$ and $\angle BCA=40^{\circ }$. We need to check whether such a triangle has the desired properties. Setting $x=y$, in isosceles triangle $BPR$ we have $\angle BRP=\angle BPR=x$. Hence, $\angle ARP=\angle BRP=x=y=\angle PCA$. This equality, along with the equations $\angle PAR=\angle CAP$ and $AP=AP$, implies that triangles $APD$ and $APC$ are congruent. Thus, $AR=AC$, or $AB+BP=AQ+BQ$. Therefore, $\angle ABC=80^{\circ }$, $\angle BCA=40^{\circ }$, and $\angle CAB=60^{\circ }$ are possible angles of triangle $ABC$.

ii. $C\ne S$.



Since $AR=AS$ and $AP$ is the angle bisector of angles $RAS$ and $BAC$, $R$ and $S$ are symmetric with respect to line $AP$, and $\angle BRD=\angle DSQ$. By (7), we have $$\angle BRD=\angle DSQ=\angle BSQ=\angle QBS=\angle QBD.(8)$$ By subtraction, from (6) and (8) we obtain $$\begin{array}{rl}\angle PRD& =\angle BRD-\angle BRP\\ & =\angle QBD-\angle QBP=\angle PBD.\end{array}$$ Because $P,D$ are on the same side of line $BR$ (specifically, between rays $AR$ and $CS$), quadrilateral $BPDR$ must be cyclic. Hence, $\angle BRD=\angle BPA$. Because angle $BPA$ is an exterior angle of triangle $APC$, by using (1), we have $\angle BPA=30^{\circ }+y=150^{\circ }-2x$. On the other hand, by (8) and (7), $\angle BRD=\angle DSQ=\angle BSQ=$ $$60^{\circ }-\frac{x}{2}.$$ Now the relation $\angle BRD=\angle BPA$ reads $$150^{\circ }-2x=60^{\circ }-\frac{x}{2},$$ or $x=60^{\circ }$. But then $y=0^{\circ }$, which is impossible. Thus, there is no solution under the assumption $C\ne S$.

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Third Solution. (By Liang Xiao, China) Construct $R$ on ray $AB$ beyond $B$ such that $BR=BP$ and $S$ on ray $AQ$ beyond $Q$ such that $QS=QB$. Note that $AR=AB+BP=AQ+QB=AS$ and hence, because $\angle SAR=60^{\circ }$, triangle $ARS$ is equilateral. Now we calculate a few angles: $$\begin{array}{rl}\angle BPA& =150^{\circ }-2x,\\ \angle BQA& =120^{\circ }-x,\\ \angle BSQ& =\angle QBS=\frac{\angle BQA}{2}=60^{\circ }-\frac{x}{2},\\ \angle ABS& =\angle ABQ+\angle QBS=60^{\circ }+\frac{x}{2},\\ \angle RSB& =60^{\circ }-\angle BSQ=\frac{x}{2},\\ \angle SBR& =180^{\circ }-\angle ABS=120^{\circ }-\frac{x}{2}.\end{array}$$ Noting that $BR=BP$ and $AS=RS$, we apply the Law of Sines to triangles $BRS$, $ABP$, and $ABS$ to obtain $$\begin{array}{rl}\frac{\sin \angle RSB}{\sin \angle SBR}& =\frac{BR}{RS}=\frac{BP}{AS}=\frac{BP}{AB}\cdot \frac{AB}{AS}\\ & =\frac{\sin \angle PAB}{\sin \angle BPA}\cdot \frac{\sin \angle BSA}{\sin \angle ABS}.\end{array}$$ Writing these angles in terms of $x$, we have $$\frac{\sin \left(\frac{x}{2}\right)}{\sin \left(120^{\circ }-\frac{x}{2}\right)}=\frac{\sin 30^{\circ }}{\sin (150^{\circ }-2x)}\cdot \frac{\sin \left(60^{\circ }-\frac{x}{2}\right)}{\sin \left(60^{\circ }+\frac{x}{2}\right)}.$$ Now, $x=(120^{\circ }-y)/2<60^{\circ }$ and hence $\sin (120^{\circ }-\frac{x}{2})=\sin (60^{\circ }+\frac{x}{2})\ne 0$. Cancelling the common terms and clearing the denominators of the above equation, we obtain $$2\sin \left(\frac{x}{2}\right)\sin (30^{\circ }+2x)=\cos \left(30^{\circ }+\frac{x}{2}\right).$$ Using the Product-to-sum formulas gives $$\cos \left(30^{\circ }+\frac{3x}{2}\right)-\cos \left(30^{\circ }+\frac{5x}{2}\right)=\cos \left(30^{\circ }+\frac{x}{2}\right),$$ or $$\cos \left(30^{\circ }+\frac{3x}{2}\right)=\cos \left(30^{\circ }+\frac{5x}{2}\right)+\cos \left(30^{\circ }+\frac{x}{2}\right).$$ Applying the Sum-to-product formulas yields $$\cos \left(30^{\circ }+\frac{3x}{2}\right)=2\cos \left(30^{\circ }+\frac{3x}{2}\right)\cos x.$$ Hence, either $\cos x=\frac{1}{2}$ or $\cos (30^{\circ }+\frac{3x}{2})=0$. The first case is impossible because $0^{\circ }<x<60^{\circ }$. In the second, $0^{\circ }<30^{\circ }+\frac{3x}{2}<120^{\circ }$, implying that $30^{\circ }+\frac{3x}{2}=90^{\circ }$ and $x=40^{\circ }$. Consequently, it is easy to check that $\angle ABS=\angle ABC=80^{\circ }$ and $S=C$. Therefore, $\angle ABC=80^{\circ }$, $\angle BCA=40^{\circ }$, and $\angle CAB=60^{\circ }$.

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Fourth Solution. (By Hui Zheng, China) Applying the Law of Sines to triangles $ABP$ and $ABQ$ yields $$\frac{AB+BP}{AB}=1+\frac{\sin 30^{\circ }}{\sin (2x+30^{\circ })}$$ and $$\frac{AQ+QB}{AB}=\frac{\sin x+\sin 60^{\circ }}{\sin (x+60^{\circ })}.$$ Thus, the condition $AB+BP=AQ+BQ$ implies that $$1+\frac{\sin 30^{\circ }}{\sin (2x+30^{\circ })}=\frac{\sin x+\sin 60^{\circ }}{\sin (x+60^{\circ })},(12)$$ or $$\frac{\sin (x+60^{\circ })}{2\sin (2x+30^{\circ })}=\sin x+\sin 60^{\circ }-\sin (x+60^{\circ }).$$ By (1), $y>0^{\circ }$ and $0^{\circ }<x<60^{\circ }$. Applying the Difference-to-product formulas twice yields $$\begin{array}{rl}\frac{\sin (x+60^{\circ })}{2\sin (2x+30^{\circ })}& =\sin 60^{\circ }-[\sin (x+60^{\circ })-\sin x]\\ & =\sin 60^{\circ }-2\sin 30^{\circ }\cos (x+30^{\circ })\\ & =\sin 60^{\circ }-\sin (60^{\circ }-x)\\ & =2\sin \frac{x}{2}\cos \left(60-\frac{x}{2}\right).\end{array}$$ By the Double-angle formulas, the numerator on the left-hand side is $$\sin (x+60^{\circ })=\sin (120^{\circ }-x)=2\sin \left(60^{\circ }-\frac{x}{2}\right)\cos \left(60^{\circ }-\frac{x}{2}\right).$$ Plugging in this expression and then dividing by $\cos (60^{\circ }-\frac{x}{2})$ — which is nonzero because $0^{\circ }<x<60^{\circ }$ — we find that $$\frac{\sin (60^{\circ }-\frac{x}{2})}{\sin (2x+30^{\circ })}=2\sin (\frac{x}{2}).$$ Clearing the denominator and applying the Product-to-sum formulas, the last equation becomes $$\begin{array}{rl}\sin \left(60^{\circ }-\frac{x}{2}\right)& =2\sin \left(\frac{x}{2}\right)\sin (2x+30^{\circ })\\ & =2\sin \left(\frac{x}{2}\right)\cos (2x-60^{\circ })\\ & =\sin \left(\frac{5x}{2}-60^{\circ }\right)+\sin \left(60^{\circ }-\frac{3x}{2}\right),\end{array}$$ or $$\begin{array}{rl}\sin \left(60-\frac{3x}{2}\right)& =\sin \left(60^{\circ }-\frac{x}{2}\right)-\sin \left(\frac{5x}{2}-60^{\circ }\right)\\ & =2\sin \left(60^{\circ }-\frac{3x}{2}\right)\cos x\end{array}$$ by applying the Difference-to-product formulas. Hence, $$\sin (60^{\circ }-\frac{3x}{2})=0\text{or}\cos x=\frac{1}{2}.$$ The only solution to either equation in the range $0^{\circ }<x<60^{\circ }$ is $x=40^{\circ }$. Thus, $\angle ABC=80^{\circ }$, $\angle BCA=40^{\circ }$, and $\angle CAB=60^{\circ }$.

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Fifth Solution. (By Kiran Kedlaya) We use complex numbers in this solution. Rewrite (12) as $$1+\frac{1}{2\cos (2x-60^{\circ })}-\frac{\sin x+\sin 60^{\circ }}{\sin (x+60^{\circ })}=0.(13)$$ Set $z=\cos x+i\sin x$ and $\omega =e^{2\pi i/6}=\cos 60^{\circ }+i\sin 60^{\circ }$. Because $0^{\circ }<x<60^{\circ }$, we know that $-1,1\ne z\omega$. Now, $\bar{z}=\frac{1}{z}$ and $\bar{\omega }=\frac{1}{\omega }$. Also, $${\omega }^3=-1\text{and}\omega -{\omega }^2=1.(14)$$ It follows that $2i\sin x=z-\bar{z}=z-\frac{1}{z}$, $2i\sin 60^{\circ }=\omega -\frac{1}{\omega }$, $2i\sin (x+60^{\circ })=z\omega -\frac{1}{z\omega }$, and $2\cos (2x-60^{\circ })=\frac{z^2}{\omega }+\frac{\omega }{z^2}$. We rewrite (13) in terms of $z$: $$1+\frac{1}{\frac{z^2}{\omega }+\frac{\omega }{z^2}}-\frac{z-\frac{1}{z}+\omega -\frac{1}{\omega }}{z\omega -\frac{1}{z\omega }}=0.(15)$$ The second term on the left-hand side of (15) is $\frac{z^2\omega }{z^4+{\omega }^2}$ and the third term simplifies to $$\frac{(z+\omega )-\frac{z+\omega }{z\omega }}{\frac{(zw{)}^2-1}{z\omega }}=\frac{(z+\omega )(z\omega -1)}{(z\omega +1)(z\omega -1)}=\frac{z+\omega }{z\omega +1}.$$ Thus, (15) simplifies to $$1+\frac{z^2\omega }{z^4+{\omega }^2}-\frac{z+\omega }{z\omega +1}=0.$$ Clearing the denominators in this equations yields $$(z^4+{\omega }^2)(z\omega +1)+z^2\omega (z\omega +1)-(z+\omega )(z^4+{\omega }^2)=0.$$ If $x\approx 60^{\circ }$, then triangles $ABQ$ and $PBQ$ are approximately equilateral and $AB+BP\approx AQ+QB$. Thus, we suspect that $z=\omega$ is a solution

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of the above equation. With this in mind, we factor (14) as follows: $$\begin{array}{rl}0& =(z^4+{\omega }^2)(z\omega +1)+z^2\omega (z\omega +1)-(z+\omega )(z^4+{\omega }^2)\\ & =z^5\omega +z{\omega }^3+z^4+{\omega }^2+z^3{\omega }^2+z^2\omega -z^5-z{\omega }^2-z^4\omega -{\omega }^3\\ & =(\omega -1)z^5+(1-\omega )z^4+{\omega }^2{z}^3+\omega z^2+({\omega }^3-{\omega }^2)z+({\omega }^2-{\omega }^3)\\ & ={\omega }^2{z}^5-{\omega }^2{z}^4+{\omega }^2{z}^3+\omega z^2-\omega z+\omega \\ & ={\omega }^2{z}^3(z^2-z+1)+\omega (z^2-z+1)\\ & ={\omega }^2(z^3-{\omega }^2)(z^2-z+1).\end{array}$$ Hence, $z^3={\omega }^2=\cos 120^{\circ }+i\sin 120^{\circ }$, so $z=\cos 40^{\circ }+i\sin 40^{\circ }$, $\cos 160^{\circ }+i\sin 160^{\circ }$, or $\cos 280^{\circ }+i\sin 280^{\circ }$. yielding $x=40^{\circ },160^{\circ }$, or $280^{\circ }$. The only value in the range $(0^{\circ },60^{\circ })$ is $x=40^{\circ }$, so $\angle ABC=80^{\circ }$, $\angle BCA=40^{\circ }$, and $\angle CAB=60^{\circ }$.

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Sixth Solution. (By Belur Jana Venkatachala, India) Let $a=BC$, $b=CA$, $c=AB$, and $2s=a+b+c$. By the Angle Bisector Theorem, we obtain $$BP=\frac{ca}{b+c}\text{and}AQ=\frac{cb}{c+a}.(16)$$ Lemma 2. We have $$BQ=\frac{2ca}{c+a}\sqrt{\frac{s(s-b)}{ca}}.$$ Proof. Let $[R]$ denote the area of region $R$. Then, $$\begin{array}{rl}2ac\sin x\cos x& =ac\sin 2x=2[ABC]=2[ABQ]+2[QBC]\\ & =c\cdot BQ\sin x+a\cdot BQ\sin x\\ & =BQ\cdot (a+c)\sin x,\end{array}$$ or $$BQ=\frac{2ac}{a+c}\cos x.(17)$$ By the Law of Cosines, we obtain $$\cos 2x=\cos \angle ABC=\frac{a^2+c^2-b^2}{2ac}.$$ Applying the Double-angle formulas yields $$\begin{array}{rl}2{\cos }^{2}x& =\cos 2x+1=\frac{a^2+c^2-b^2}{2ac}+1\\ & =\frac{(a^2+2ac+c^2)-b^2}{2ac}=\frac{(a+c{)}^2-b^2}{2ac}\\ & =\frac{(a+c+b)(a+c-b)}{2ac}=\frac{2s(s-b)}{ac},\end{array}$$ or $\cos x=\sqrt{\frac{s(s-b)}{ac}}$. Substituting this into (17) yields the desired result. (The result also follows directly from the Angle Bisector Theorem and the Stewart's Theorem.) ■

By lemma 2, the given condition reads $$c+\frac{ca}{b+c}=\frac{cb}{c+a}+\frac{2ca}{c+a}\sqrt{\frac{s(s-b)}{ca}},$$ or $$\frac{a+b+c}{b+c}=\frac{b}{a+c}+\frac{2a}{c+a}\sqrt{\frac{s(s-b)}{ca}}.$$ Clearing the denominators yields $$(a+b+c)(c+a)\sqrt{c}=b(b+c)\sqrt{c}+2\sqrt{a}(b+c)\sqrt{s(s-b)}.(18)$$ Since $$\begin{array}{rl}(a+b+c)(c+a)-b(b+c)& =(a+b+c)(c+a)-b(a+b+c)+ab\\ & =(a+b+c)(a+c-b)+ab\\ & =4s(s-b)+ab,\end{array}$$ (18) can be rewritten as $$4\sqrt{cs(s-b)}+ab\sqrt{c}=2\sqrt{a}(b+c)\sqrt{s(s-b)}.$$ Taking $t=\sqrt{s(s-b)}$, we obtain a quadratic in $t$: $$4\sqrt{c{t}^2-2\sqrt{a}(b+c)t+ab\sqrt{c}}=0,$$ or $(2t-\sqrt{ac})(2\sqrt{ct}-\sqrt{ab})=0$. Thus, $$t=\frac{\sqrt{ac}}{2}\text{or}t=\frac{\sqrt{ab}}{2\sqrt{c}}.$$ If $t=\frac{\sqrt{ac}}{2}$, then $s(s-b)=\frac{ac}{4}$, leading to $$\frac{(a+c+b)(a+c-b)}{4}=\frac{(a+c{)}^2-b^2}{4}=\frac{ac}{4},$$ or $a^2+ac+c^2=b^2$. This equation and the Law of Cosines imply that $\angle ABC=120^{\circ }$, which is absurd.

If $t=\frac{\sqrt{ab}}{2\sqrt{c}}$, then $s(s-b)=\frac{a{b}^2}{4c}$, leading to $$(a+c{)}^2-b^2=\frac{a{b}^2}{c}.$$ Hence, $$(a+c{)}^2=\frac{(a+c)b^2}{c},$$ or $$\frac{c}{b}=\frac{cb}{a+c}\cdot \frac{1}{c}.$$ By the second part of (16), the last equation is $$\frac{AB}{AC}=\frac{AQ}{AB},$$ which, along with $\angle BAC=\angle QAB$, implies that triangles $ABQ$ and $ACB$ are similar. Hence, $\angle ACB=\angle ABQ=\angle BQC$, that is, $\angle ABC=80^{\circ }$, $\angle BCA=40^{\circ }$, and $\angle CAB=60^{\circ }$.

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Therefore, $\angle ABC=80^{\circ }$, $\angle BCA=40^{\circ }$, and $\angle CAB=60^{\circ }$ is the only solution.