United States — Team Selection Test, 2010

Solution 1. We begin with a key lemma, for which we provide two proofs.
Lemma 1. Let $a = BC$, $b = CA$, $c = AB$, and let $p_a$, $p_b$, $p_c$ denote the distances from $P$ to sides $BC$, $CA$, $AB$, respectively. We have
$$2a \cdot AP \geq (b+c)(p_b+p_c). \qquad (1)$$

First proof of Lemma 1. First, notice that $\frac{p_a}{h_a} = \frac{[PBC]}{[ABC]}$, $\frac{p_b}{h_b} = \frac{[PCA]}{[BCA]}$, and $\frac{p_c}{h_c} = \frac{[PAB]}{[CAB]}$, hence we have
$$\frac{p_a}{h_a} + \frac{p_b}{h_b} + \frac{p_c}{h_c} = 1. \qquad (2)$$
Let $E$ and $D$ be the feet of the perpendiculars from $P$ to sides $AB$ and $AC$, respectively. (Thus, $PD = p_b$ and $PE = p_c$.) Let points $M$ and $N$ lie on sides $AB$ and $AC$, respectively, such that $MN \parallel BC$ and $P$ lies on $MN$. We have $MN \cdot AP \ge 2[AMN]$ and $2[AMN] = 2[APM] + 2[APN] = PE \cdot AM + PD \cdot AN$. Hence
$$MN \cdot AP \ge PE \cdot AM + PD \cdot AN.$$
Because $AMN$ and $ABC$ are similar, we have $MN : AM : AN = BC : AB : AC = a : c : b$, hence the last inequality implies that
$$a \cdot AP \ge bp_b + cp_c. \qquad (3)$$
Let $P_1$ be the reflection of $P$ across the bisector of $\angle CAB$, and let $E_1$ and $D_1$ be the feet of the perpendiculars from $P_1$ to sides $AB$ and $AC$, respectively. By symmetry, it is not difficult to see that $AE_1 = AD$, $AD_1 = AE$, $P_1E_1 = PD = p_b$, and $P_1D_1 = PE = p_c$. In exactly the same way we established (3), we can show that
$$a \cdot AP_1 \ge bp_c + cp_b. \qquad (4)$$
Adding (3) and (4) together yields
$$2a \cdot AP = a \cdot AP + a \cdot AP_1 \ge bp_b + cp_c + bp_c + cp_b = (b+c)(p_b + p_c),$$
establishing the lemma. $\square$

Second proof of Lemma 1. Set $\angle CAB = A$, $\angle ABC = B$, $\angle BCA = C$, $\angle BAP = x$ and $\angle CAP = y$. Note that (1) is equivalent to
$$\frac{2PA}{p_b + p_c} \ge \frac{b+c}{a}.$$
Further, we have that
$$\frac{2PA}{p_b + p_c} = \frac{2}{\frac{p_b}{PA} + \frac{p_c}{PA}} = \frac{2}{\sin \alpha + \sin \beta}$$
and $\frac{b+c}{a} = \frac{\sin B + \sin C}{\sin A}$ by the law of sines, so it suffices to show that
$$2 \sin A \ge (\sin \alpha + \sin \beta)(\sin B + \sin C).$$
Indeed, by the sum-to-product formulas, we obtain
$$\begin{align*} (\sin \alpha + \sin \beta)(\sin B + \sin C) &= 4 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \sin \frac{B+C}{2} \cos \frac{B-C}{2} \\ &\le 4 \sin \frac{\alpha + \beta}{2} \sin \frac{B+C}{2} \\ &= 4 \sin \frac{A}{2} \cos \frac{A}{2} \\ &= 2 \sin A, \end{align*}$$
as needed. $\square$

We now proceed to the problem proper. By Lemma 1, we have
$$\begin{aligned} \frac{PA}{h_b + h_c} &\ge \frac{(b+c)(p_b + p_c)}{2a(h_b + h_c)} = \frac{(b+c)(p_b + p_c)}{2a\left(\frac{2[ABC]}{b} + \frac{2[ABC]}{c}\right)} = \frac{bc(p_b + p_c)}{4a[ABC]} \\ &= \frac{c(bp_b)}{4a[ABC]} + \frac{b(cpc)}{4a[ABC]} = \frac{c[PCA]}{2a[ABC]} + \frac{b[PAB]}{2a[ABC]} = \frac{c}{2a} \cdot \frac{p_b}{h_b} + \frac{b}{2a} \cdot \frac{p_c}{h_c}. \end{aligned}$$
Likewise, we have
$$\frac{PB}{h_c + h_a} \ge \frac{a}{2b} \cdot \frac{p_c}{h_c} + \frac{c}{2b} \cdot \frac{p_a}{h_a} \quad \text{and} \quad \frac{PC}{h_a + h_b} \ge \frac{b}{2c} \cdot \frac{p_a}{h_a} + \frac{a}{2c} \cdot \frac{p_b}{h_b}.$$
Adding the last three inequalities gives
$$\begin{aligned} \frac{PA}{h_b + h_c} + \frac{PB}{h_c + h_a} + \frac{PC}{h_a + h_b} &\ge \frac{c}{2a} \cdot \frac{p_b}{h_b} + \frac{b}{2a} \cdot \frac{p_c}{h_c} + \frac{a}{2b} \cdot \frac{p_c}{h_c} + \frac{c}{2b} \cdot \frac{p_a}{h_a} + \frac{b}{2c} \cdot \frac{p_a}{h_a} + \frac{a}{2c} \cdot \frac{p_b}{h_b} \\ &= \left( \frac{c}{2b} + \frac{b}{2c} \right) \frac{p_a}{h_a} + \left( \frac{a}{2c} + \frac{c}{2a} \right) \frac{p_b}{h_b} + \left( \frac{b}{2a} + \frac{a}{2b} \right) \frac{p_c}{h_c} \\ &\ge \frac{p_a}{h_a} + \frac{p_b}{h_b} + \frac{p_c}{h_c} \\ &= 1 \end{aligned}$$
by the AM-GM inequality and (2).

Solution 2 (by Gabriel Carroll). We start with the following observation.
Lemma 2. Let $XYZ$ be a triangle, and let $\omega_X$ be the excircle opposite $X$. Let $O$ and $R$ be the center and radius of $\omega_X$, respectively. Then
$$d(Y, XZ) + d(Z, XY) \le XO.$$
(For a point $P$ and a line $\ell$, $d(P, \ell)$ denotes the distance from $P$ to $\ell$.)
Proof. Fix the excircle and the two tangent rays from $X$, and let $Y$ and $Z$ be variable points on these rays such that $YZ$ is tangent to the circle, and the circle lies on the opposite side of line $YZ$ from $X$. Let the tangent ray from $X$ passing through $Y$ touch the excircle at $S$ (and so $OS = R$). Then we have
$$d(Y, XZ) + d(Z, XY) = (XY + XZ) \sin \angle ZXY = (2SX - YZ) \sin \angle ZXY,$$
which is maximized by minimizing $YZ$. Let $YZ$ be tangent to the circle at $T$. As $Y$ and $Z$ vary, $\angle YOZ$ and $\angle YOT + \angle TOZ = 180^\circ - \angle YOZ$ are both constants, and
$$YZ = R(\tan \angle YOT + \tan \angle TOZ).$$
By the convexity of the tangent function, $YZ$ is minimized when $\angle YOT = \angle TOZ$; that is, $YZ \perp XO$ or $XYZ$ is isosceles. Hence it suffices to prove our result in this case. We may set $\angle XOS = 2\gamma$. Then $\angle SOY = \gamma$, $\angle OXY = \angle OXS = 90^\circ - 2\gamma$, $\angle ZXY = 2\angle OXY = 180^\circ - 4\gamma$, $XO = R\sec 2\gamma$, $XS = R\tan 2\gamma$, $YS = R\tan \gamma$, $XY = R(\tan 2\gamma - \tan \gamma)$, and $d(Y, XZ) = XY \sin \angle ZXY = R(\tan 2\gamma - \tan \gamma) \sin 4\gamma$. We need to show that
$$d(Y, XZ) + d(Z, XY) = 2d(Y, XZ) \le XO \quad \text{or} \quad 2(\tan 2\gamma - \tan \gamma) \sin 4\gamma \le \sec 2\gamma.$$
Now, notice that $\gamma < 45^\circ$, so we have $\cos 2\gamma \ge 0$ and $\cos \gamma \ge 0$. Thus, multiplying both sides of the last desired inequality by $\cos 2\gamma \cos \gamma$, it suffices to show that
$$2 \sin 4\gamma (\sin 2\gamma \cos \gamma - \cos 2\gamma \sin \gamma) \le \cos \gamma,$$
or equivalently by the sine subtraction formula that $2 \sin 4\gamma \sin \gamma \le \cos \gamma$. By the double-angle formula, the last inequality is equivalent to $8 \cos 2\gamma \cos \gamma \sin^2 \gamma \le \cos \gamma$, or $8(1 - 2 \sin^2 \gamma) \sin^2 \gamma \le 1$. The latter is equivalent to $(4 \sin^2 \gamma - 1)^2 \ge 0$, establishing the result. $\square$

We now return to the original problem. Let $A_1$, $B_1$, $C_1$ be the excenters of $ABC$ opposite $A$, $B$, $C$. It is well known that $A_1A$, $B_1B$, and $C_1C$ are the altitudes of triangle $A_1B_1C_1$. By Lemma 2, $h_b + h_c \le A_1A$, implying that
$$\frac{PA}{h_b + h_c} \ge \frac{PA}{A_1A} \ge \frac{[PB_1C_1]}{[A_1B_1C_1]}.$$
Adding the last inequality and its cyclic analogs yields
$$\frac{PA}{h_b + h_c} + \frac{PB}{h_c + h_a} + \frac{PC}{h_a + h_b} \ge \frac{[PB_1C_1] + [PC_1A_1] + [PA_1B_1]}{[A_1B_1C_1]} = 1.$$