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Difference between revisions of "2001 USAMO Problems/Problem 4"
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== Solution ==
 
== Solution ==
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{{solution}}
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=== Solution 1 ===
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We know that <math>PB^2+PC^2 < PA^2</math> and we wish to prove that <math>AB^2 + AC^2 > BC^2</math>.
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It would be sufficient to prove that
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<cmath>PB^2+PC^2+AB^2+AC^2 \geq PA^2 + BC^2.</cmath>
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Set <math>A(0,0)</math>, <math>B(1,0)</math>, <math>C(x,y)</math>, <math>P(p,q)</math>.
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Then, we wish to show
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<cmath>(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 </cmath>
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<cmath>2p^2 + 2q^2 + 2x^2 + 2y^2 - 2p - 2px - 2qy + 2 \geq p^2 + q^2 + x^2 + y^2 - 2x + 1 </cmath>
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<cmath>p^2 + q^2 + x^2 + y^2 + 2x - 2p - 2px - 2qy + 1 \geq 0 </cmath>
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<cmath>(x-p)^2 + (q-y)^2 + 2(x-p) + 1 \geq 0 </cmath>
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<cmath>(x-p+1)^2 + (q-y)^2 \geq 0,</cmath>
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which is true by the trivial inequality.
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=== Solution 2 ===
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Let <math>A</math> be the origin. For a point <math>Q</math>, denote by <math>q</math> the vector <math>\overrightarrow{AQ}</math>, and denote by <math>|q|</math> the length of <math>q</math>. The given conditions may be written as
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<cmath>|p - b|^2 + |p - c|^2 < |p|^2,</cmath>
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or
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<cmath>p\cdot p + b\cdot b + c\cdot c - 2p\cdot b - 2p\cdot c < 0.</cmath>
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Adding <math>2b\cdot c</math> on both sides of the last inequality gives
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<cmath>|p - b - c|^2 < 2b\cdot c.</cmath>
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Since the left-hand side of the last inequality is nonnegative, the right-hand side is positive. Hence
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<cmath>\cos\angle BAC = \frac{b\cdot c}{|b||c|} > 0,</cmath>
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that is, <math>\angle BAC</math> is acute.
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=== Solution 3 ===
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For the sake of contradiction, let's assume to the contrary that <math>\angle BAC</math>. Let <math>AB = c</math>, <math>BC = a</math>, and <math>CA = b</math>. Then <math>a^2\geq b^2 + c^2</math>. We claim that the quadrilateral <math>ABPC</math> is convex. Now applying the generalized Ptolemy's Theorem to the convex quadrilateral <math>ABPC</math> yields
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<cmath>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},</cmath>
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where the second inequality is by Cauchy-Schwarz. This implies <math>PA^2\leq PB^2 + PC^2</math>, in contradiction with the facts that <math>PA</math>, <math>PB</math>, and <math>PC</math> are the sides of an obtuse triangle and <math>PA > \max\{PB, PC\}</math>.
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We present two arguments to prove our claim.
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''First argument'': Without loss of generality, we may assume that <math>A</math>, <math>B</math>, and <math>C</math> are in counterclockwise order. Let lines <math>l_1</math> and <math>l_2</math> be the perpendicular bisectors of segments <math>AB</math> and <math>AC</math>, respectively. Then <math>l_1</math> and <math>l_2</math> meet at <math>O</math>, the circumcenter of triangle <math>ABC</math>. Lines <math>l_1</math> and <math>l_2</math> cut the plane into four regions and <math>A</math> is in the interior of one of these regions. Since <math>PA > PB</math> and <math>PA > PC</math>, <math>P</math> must be in the interior of the region that opposes <math>A</math>. Since <math>\angle BAC</math> is not acute, ray <math>AC</math> does not meet <math>l_1</math> and ray <math>AB</math> does not meet <math>l_2</math>. Hence <math>B</math> and <math>C</math> must lie in the interiors of the regions adjacent to <math>A</math>. Let <math>\mathcal{R}_X</math> denote the region containing <math>X</math>. Then <math>\mathcal{R}_A</math>, <math>\mathcal{R}_B</math>, <math>\mathcal{R}_P</math>, and <math>\mathcal{R}_C</math> are the four regions in counterclockwise order. Since <math>\angle BAC\geq 90^\circ</math>, either <math>O</math> is on side <math>BC</math> or <math>O</math> and <math>A</math> are on opposite sides of line <math>BC</math>. In either case <math>P</math> and <math>A</math> are on opposite sides of line <math>BC</math>. Also, since ray <math>AB</math> does not meet <math>l_2</math> and ray <math>AC</math> does not meet <math>l_1</math>, it follows that <math>\mathcal{R}_P</math> is entirely in the interior of <math>\angle BAC</math>. Hence <math>B</math> and <math>C</math> are on opposite sides of <math>AP</math>. Therefore <math>ABPC</math> is convex.
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<center>[[File:2001usamo4-1.png]]</center>
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''Second argument'': Since <math>PA > PB</math> and <math>PA > PC</math>, <math>A</math> cannot be inside or on the sides of triangle <math>PBC</math>. Since <math>PA > PB</math>, we have <math>\angle ABP > \angle BAP</math> and hence <math>\angle BAC\geq 90^\circ > \angle BAP</math>. Hence <math>C</math> cannot be inside or on the sides of triangle <math>BAP</math>. Symmetrically, <math>B</math> cannot be inside or on the sides of triangle <math>CAP</math>. Finally, since <math>\angle ABP > \angle BAP</math> and <math>\angle ACP > \angle CAP</math>, we have
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<cmath>\angle ABP + \angle ACP > \angle BAC\geq 90^\circ\geq\angle ABC + \angle ACB.</cmath>
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Therefore <math>P</math> cannot be inside or on the sides of triangle <math>ABC</math>. Since this covers all four cases, <math>ABPC</math> is convex.
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==Solution 4==
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Let <math>P</math> be the origin in vector space, and let <math>a, b, c</math> denote the position vectors of <math>A, B, C</math> respectively. Then the obtuse triangle condition, <math>PA^2 > PB^2 + PC^2</math>, becomes <math>a^2 > b^2 + c^2</math> 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 <math>\angle{BAC}</math> is acute, it suffices to show that <math>(a - b)(a - c) > 0</math>, or <math>a^2 - ab - ac + bc > 0</math>. But this follows from the observation that
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<cmath>(-a + b + c)^2 \ge 0,</cmath>
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which leads to
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<cmath>2a^2 - 2ab - 2ac + 2bc > a^2 + b^2 + c^2 - 2ab - 2ac + 2bc \ge 0</cmath>
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and therefore our desired conclusion.
  
 
== See also ==
 
== See also ==
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[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
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{{MAA Notice}}

Revision as of 13:08, 10 March 2016

Problem

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

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.

2001usamo4-1.png

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.

See also

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

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