2001 USAMO Problems/Problem 4

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Let $P$ be a point in the plane of triangle $ABC$ such that the segments $PA$, $PB$, and $PC$ are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to $PA$. Prove that $\angle BAC$ is acute.


Solution 1

We know that $PB^2+PC^2 < PA^2$ and we wish to prove that $AB^2 + AC^2 > BC^2$. It would be sufficient to prove that \[PB^2+PC^2+AB^2+AC^2 \geq PA^2 + BC^2.\] Set $A(0,0)$, $B(1,0)$, $C(x,y)$, $P(p,q)$. Then, we wish to show

\[(p-1)^2 + q^2 + (p-x)^2 + (q-y)^2 + 1 + x^2 + y^2 \geq p^2 + q^2 + (x-1)^2 + y^2\] \[2p^2 + 2q^2 + 2x^2 + 2y^2 - 2p - 2px - 2qy + 2 \geq p^2 + q^2 + x^2 + y^2 - 2x + 1\] \[p^2 + q^2 + x^2 + y^2 + 2x - 2p - 2px - 2qy + 1 \geq 0\] \[(x-p)^2 + (q-y)^2 + 2(x-p) + 1 \geq 0\] \[(x-p+1)^2 + (q-y)^2 \geq 0,\]

which is true by the trivial inequality.

Solution 2

Let $A$ be the origin. For a point $Q$, denote by $q$ the vector $\overrightarrow{AQ}$, and denote by $|q|$ the length of $q$. The given conditions may be written as \[|p - b|^2 + |p - c|^2 < |p|^2,\] or \[p\cdot p + b\cdot b + c\cdot c - 2p\cdot b - 2p\cdot c < 0.\] Adding $2b\cdot c$ on both sides of the last inequality gives \[|p - b - c|^2 < 2b\cdot c.\] Since the left-hand side of the last inequality is nonnegative, the right-hand side is positive. Hence \[\cos\angle BAC = \frac{b\cdot c}{|b||c|} > 0,\] that is, $\angle BAC$ is acute.

Solution 3

For the sake of contradiction, let's assume to the contrary that $\angle BAC$. Let $AB = c$, $BC = a$, and $CA = b$. Then $a^2\geq b^2 + c^2$. We claim that the quadrilateral $ABPC$ is convex. Now applying the generalized Ptolemy's Theorem to the convex quadrilateral $ABPC$ yields \[a\cdot PA\leq b\cdot PB + c\cdot PC\leq\sqrt{b^2 + c^2}\sqrt{PB^2 + PC^2}\leq a\sqrt{PB^2 + PC^2},\] where the second inequality is by Cauchy-Schwarz. This implies $PA^2\leq PB^2 + PC^2$, in contradiction with the facts that $PA$, $PB$, and $PC$ are the sides of an obtuse triangle and $PA > \max\{PB, PC\}$.

We present two arguments to prove our claim.

First argument: Without loss of generality, we may assume that $A$, $B$, and $C$ are in counterclockwise order. Let lines $l_1$ and $l_2$ be the perpendicular bisectors of segments $AB$ and $AC$, respectively. Then $l_1$ and $l_2$ meet at $O$, the circumcenter of triangle $ABC$. Lines $l_1$ and $l_2$ cut the plane into four regions and $A$ is in the interior of one of these regions. Since $PA > PB$ and $PA > PC$, $P$ must be in the interior of the region that opposes $A$. Since $\angle BAC$ is not acute, ray $AC$ does not meet $l_1$ and ray $AB$ does not meet $l_2$. Hence $B$ and $C$ must lie in the interiors of the regions adjacent to $A$. Let $\mathcal{R}_X$ denote the region containing $X$. Then $\mathcal{R}_A$, $\mathcal{R}_B$, $\mathcal{R}_P$, and $\mathcal{R}_C$ are the four regions in counterclockwise order. Since $\angle BAC\geq 90^\circ$, either $O$ is on side $BC$ or $O$ and $A$ are on opposite sides of line $BC$. In either case $P$ and $A$ are on opposite sides of line $BC$. Also, since ray $AB$ does not meet $l_2$ and ray $AC$ does not meet $l_1$, it follows that $\mathcal{R}_P$ is entirely in the interior of $\angle BAC$. Hence $B$ and $C$ are on opposite sides of $AP$. Therefore $ABPC$ is convex.


Second argument: Since $PA > PB$ and $PA > PC$, $A$ cannot be inside or on the sides of triangle $PBC$. Since $PA > PB$, we have $\angle ABP > \angle BAP$ and hence $\angle BAC\geq 90^\circ > \angle BAP$. Hence $C$ cannot be inside or on the sides of triangle $BAP$. Symmetrically, $B$ cannot be inside or on the sides of triangle $CAP$. Finally, since $\angle ABP > \angle BAP$ and $\angle ACP > \angle CAP$, we have \[\angle ABP + \angle ACP > \angle BAC\geq 90^\circ\geq\angle ABC + \angle ACB.\] Therefore $P$ cannot be inside or on the sides of triangle $ABC$. Since this covers all four cases, $ABPC$ is convex.

Solution 4

Let $P$ be the origin in vector space, and let $a, b, c$ denote the position vectors of $A, B, C$ respectively. Then the obtuse triangle condition, $PA^2 > PB^2 + PC^2$, becomes $a^2 > b^2 + c^2$ using the fact that the square of a vector (the dot product of itself and itself) is the square of its magnitude. Now, notice that to prove $\angle{BAC}$ is acute, it suffices to show that $(a - b)(a - c) > 0$, or $a^2 - ab - ac + bc > 0$. But this follows from the observation that \[(-a + b + c)^2 \ge 0,\] which leads to \[2a^2 - 2ab - 2ac + 2bc > a^2 + b^2 + c^2 - 2ab - 2ac + 2bc \ge 0\] and therefore our desired conclusion.

Solution 5

Let $M, N$ be midpoints of $AP$ and $BC$, respectively. For the points $A, B, P, C$; let's apply Euler's quadrilateral formula, \[AB^2 + BP^2 + PC^2 + CA^2 = AP^2 + BC^2 + 4MN^2 \geq AP^2 + BC^2 .\] Given that $AP^2 > BP^2 + PC^2$. Thus, \[AB^2 + AC^2 > BC^2 .\] and we get $\angle BAC$ is acute.

(Lokman GÖKÇE)

Solution 6

Without loss of generality, assume that in a Cartesian coordinate system, $A$ is at the point $(0, 0)$ and $C$ is at the point $(1,0)$. Let $B$ be at the point $(b_x,b_y)$ and $P$ be at the point $(p_x,p_y)$. Without loss of generality, also assume that $b_y>0$.

Now, assume for contradiction that $\angle BAC$ is not acute. Since $PA$, $PB$, and $PC$ are the sides of an obtuse triangle, with $PA$ the longest side, it follows that $PA^2>PB^2+PC^2$, implying that $p_x^2+p_y^2>(p_x-b_x)^2+(p_y-b_y)^2+(p_x-1)^2+p_y^2$. This inequality simplifies to $b_x^2-2p_x b_x+b_y^2-2p_y b_y+p_x^2-2p_x+1+p_y^2<0$. Note that since $p_x^2-2p_x+1$ and $b_y^2-2p_y b_y+p_y^2$ are both perfect squares, all terms of this inequality except for $-2p_x b_x$ are already guaranteed to be nonnegative.

If $p_x<0$, then $P$ would be closer to $A$ than to $C$, but since $PA^2=PB^2+PC^2$, this is not possible. Therefore, $p_x \geq 0$. Since $\angle BAC$ not being acute implies that $b_x \leq 0$, it follows that $-2p_x b_x \geq 0$. But now since all terms of $b_x^2-2p_x b_x+b_y^2-2p_y b_y+p_x^2-2p_x+1+p_y^2<0$ are guaranteed to be nonnegative, this entire expression cannot be negative, leading to a contradiction. Therefore, $\angle BAC$ is acute.

See also

2001 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6
All USAMO Problems and Solutions

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