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
$$\angle BRP = \angle RPB = (180^\circ - \angle PBR)/2 = x = \angle QBP. \quad (2)$$
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,
$$PR = PS \qquad (3)$$
and $\angle ARP = \angle PSA$, or $\angle BRP = \angle PSQ$. By (2), we have
$$\angle QBP = \angle BRP = \angle PSQ. \qquad (4)$$

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

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
$$\angle BRP = \angle RPB = \frac{\angle ABC}{2} = x = \angle QBP \qquad (6)$$
and
$$\angle QBS = \angle BSQ = \frac{\angle BQA}{2} = \frac{180^\circ - 60^\circ - x}{2} = 60^\circ - \frac{x}{2}. \quad (7)$$

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. $C \neq S$.

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
$$\angle BRD = \angle DSQ = \angle BSQ = \angle QBS = \angle QBD. \quad (8)$$
By subtraction, from (6) and (8) we obtain
$$\begin{aligned} \angle PRD &= \angle BRD - \angle BRP \\ &= \angle QBD - \angle QBP = \angle PBD. \end{aligned}$$
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 $C \neq S$.

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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:
$$\begin{aligned} \angle BPA &= 150^\circ - 2x, \\ \angle BQA &= 120^\circ - x, \\ \angle BSQ &= \angle QBS = \frac{\angle BQA}{2} = 60^\circ - \frac{x}{2}, \\ \angle ABS &= \angle ABQ + \angle QBS = 60^\circ + \frac{x}{2}, \\ \angle RSB &= 60^\circ - \angle BSQ = \frac{x}{2}, \\ \angle SBR &= 180^\circ - \angle ABS = 120^\circ - \frac{x}{2}. \end{aligned}$$
Noting that $BR = BP$ and $AS = RS$, we apply the Law of Sines to triangles $BRS$, $ABP$, and $ABS$ to obtain
$$\begin{aligned} \frac{\sin \angle RSB}{\sin \angle SBR} &= \frac{BR}{RS} = \frac{BP}{AS} = \frac{BP}{AB} \cdot \frac{AB}{AS} \\ &= \frac{\sin \angle PAB}{\sin \angle BPA} \cdot \frac{\sin \angle BSA}{\sin \angle ABS}. \end{aligned}$$
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 $\sin(120^\circ - \frac{x}{2}) = \sin(60^\circ + \frac{x}{2}) \neq 0$. 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
$$1 + \frac{\sin 30^\circ}{\sin(2x + 30^\circ)} = \frac{\sin x + \sin 60^\circ}{\sin(x + 60^\circ)}, \quad (12)$$
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
$$\begin{aligned} \frac{\sin(x + 60^\circ)}{2 \sin(2x + 30^\circ)} &= \sin 60^\circ - [\sin(x + 60^\circ) - \sin x] \\ &= \sin 60^\circ - 2 \sin 30^\circ \cos(x + 30^\circ) \\ &= \sin 60^\circ - \sin(60^\circ - x) \\ &= 2 \sin \frac{x}{2} \cos \left(60 - \frac{x}{2}\right). \end{aligned}$$
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
$$\begin{aligned} \sin\left(60^\circ - \frac{x}{2}\right) &= 2 \sin\left(\frac{x}{2}\right) \sin(2x + 30^\circ) \\ &= 2 \sin\left(\frac{x}{2}\right) \cos(2x - 60^\circ) \\ &= \sin\left(\frac{5x}{2} - 60^\circ\right) + \sin\left(60^\circ - \frac{3x}{2}\right), \end{aligned}$$
or
$$\begin{aligned} \sin\left(60 - \frac{3x}{2}\right) &= \sin\left(60^\circ - \frac{x}{2}\right) - \sin\left(\frac{5x}{2} - 60^\circ\right) \\ &= 2 \sin\left(60^\circ - \frac{3x}{2}\right) \cos x \end{aligned}$$
by applying the Difference-to-product formulas. Hence,
$$\sin(60^\circ - \frac{3x}{2}) = 0 \quad \text{or} \quad \cos x = \frac{1}{2}.$$
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
$$1 + \frac{1}{2\cos(2x - 60^\circ)} - \frac{\sin x + \sin 60^\circ}{\sin(x + 60^\circ)} = 0. \quad (13)$$
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 $-1, 1 \neq z\omega$. Now, $\bar{z} = \frac{1}{z}$ and $\bar{\omega} = \frac{1}{\omega}$. Also,
$$\omega^3 = -1 \quad \text{and} \quad \omega - \omega^2 = 1. \quad (14)$$
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 $z$:
$$1 + \frac{1}{\frac{z^2}{\omega} + \frac{\omega}{z^2}} - \frac{z - \frac{1}{z} + \omega - \frac{1}{\omega}}{z\omega - \frac{1}{z\omega}} = 0. \quad (15)$$
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:
$$\begin{align*} 0 &= (z^4 + \omega^2)(z\omega + 1) + z^2\omega(z\omega + 1) - (z + \omega)(z^4 + \omega^2) \\ &= z^5\omega + z\omega^3 + z^4 + \omega^2 + z^3\omega^2 + z^2\omega - z^5 - z\omega^2 - z^4\omega - \omega^3 \\ &= (\omega - 1)z^5 + (1 - \omega)z^4 + \omega^2 z^3 + \omega z^2 + (\omega^3 - \omega^2)z + (\omega^2 - \omega^3) \\ &= \omega^2 z^5 - \omega^2 z^4 + \omega^2 z^3 + \omega z^2 - \omega z + \omega \\ &= \omega^2 z^3 (z^2 - z + 1) + \omega (z^2 - z + 1) \\ &= \omega^2 (z^3 - \omega^2) (z^2 - z + 1). \end{align*}$$
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
$$BP = \frac{ca}{b+c} \quad \text{and} \quad AQ = \frac{cb}{c+a}. \qquad (16)$$
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,
$$\begin{align*} 2ac \sin x \cos x &= ac \sin 2x = 2[ABC] = 2[ABQ] + 2[QBC] \\ &= c \cdot BQ \sin x + a \cdot BQ \sin x \\ &= BQ \cdot (a+c) \sin x, \end{align*}$$
or
$$BQ = \frac{2ac}{a+c} \cos x. \qquad (17)$$
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
$$\begin{aligned} 2 \cos^2 x &= \cos 2x + 1 = \frac{a^2 + c^2 - b^2}{2ac} + 1 \\ &= \frac{(a^2 + 2ac + c^2) - b^2}{2ac} = \frac{(a+c)^2 - b^2}{2ac} \\ &= \frac{(a+c+b)(a+c-b)}{2ac} = \frac{2s(s-b)}{ac}, \end{aligned}$$
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
$$(a+b+c)(c+a)\sqrt{c} = b(b+c)\sqrt{c} + 2\sqrt{a}(b+c)\sqrt{s(s-b)}. \quad (18)$$
Since
$$\begin{aligned} (a+b+c)(c+a) - b(b+c) &= (a+b+c)(c+a) - b(a+b+c) + ab \\ &= (a+b+c)(a+c-b) + ab \\ &= 4s(s-b) + ab, \end{aligned}$$
(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 $t$:
$$4\sqrt{ct^2 - 2\sqrt{a}(b+c)t + ab\sqrt{c}} = 0,$$
or $(2t - \sqrt{ac})(2\sqrt{ct} - \sqrt{ab}) = 0$. Thus,
$$t = \frac{\sqrt{ac}}{2} \quad \text{or} \quad t = \frac{\sqrt{ab}}{2\sqrt{c}}.$$
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 $s(s-b) = \frac{ab^2}{4c}$, leading to
$$(a+c)^2 - b^2 = \frac{ab^2}{c}.$$
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.