Difference between revisions of "2001 USAMO Problems/Problem 4"
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== Solution == | == Solution == | ||
− | + | === Solution 1 === | |
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>. | 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>. | ||
It would be sufficient to prove that | It would be sufficient to prove that | ||
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which is true by the trivial inequality. | which is true by the trivial inequality. | ||
+ | |||
+ | === Solution 2 === | ||
+ | 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 | ||
+ | <cmath>|p - b|^2 + |p - c|^2 < |p|^2,</cmath> | ||
+ | or | ||
+ | <cmath>p\cdot p + b\cdot b + c\cdot c - 2p\cdot b - 2p\cdot c < 0.</cmath> | ||
+ | Adding <math>2b\cdot c</math> on both sides of the last inequality gives | ||
+ | <cmath>|p - b - c|^2 < 2b\cdot c.</cmath> | ||
+ | Since the left-hand side of the last inequality is nonnegative, the right-hand side is positive. Hence | ||
+ | <cmath>\cos\angle BAC = \frac{b\cdot c}{|b||c|} > 0,</cmath> | ||
+ | that is, <math>\angle BAC</math> is acute. | ||
+ | |||
+ | === Solution 3 === | ||
+ | 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 | ||
+ | <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> | ||
+ | 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>. | ||
+ | |||
+ | We present two arguments to prove our claim. | ||
+ | |||
+ | ''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. | ||
+ | |||
+ | <center>[[File:2001usamo4-1.png]]</center> | ||
+ | |||
+ | ''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 | ||
+ | <cmath>\angle ABP + \angle ACP > \angle BAC\geq 90^\circ\geq\angle ABC + \angle ACB.</cmath> | ||
+ | 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. | ||
== See also == | == See also == |
Revision as of 12:30, 15 July 2014
Problem
Let be a point in the plane of triangle such that the segments , , and are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to . Prove that is acute.
Solution
Solution 1
We know that and we wish to prove that . It would be sufficient to prove that Set , , , . 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,
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)which is true by the trivial inequality.
Solution 2
Let be the origin. For a point , denote by the vector , and denote by the length of . The given conditions may be written as or Adding on both sides of the last inequality gives Since the left-hand side of the last inequality is nonnegative, the right-hand side is positive. Hence that is, is acute.
Solution 3
For the sake of contradiction, let's assume to the contrary that . Let , , and . Then . We claim that the quadrilateral is convex. Now applying the generalized Ptolemy's Theorem to the convex quadrilateral yields where the second inequality is by Cauchy-Schwarz. This implies , in contradiction with the facts that , , and are the sides of an obtuse triangle and .
We present two arguments to prove our claim.
First argument: Without loss of generality, we may assume that , , and are in counterclockwise order. Let lines and be the perpendicular bisectors of segments and , respectively. Then and meet at , the circumcenter of triangle . Lines and cut the plane into four regions and is in the interior of one of these regions. Since and , must be in the interior of the region that opposes . Since is not acute, ray does not meet and ray does not meet . Hence and must lie in the interiors of the regions adjacent to . Let denote the region containing . Then , , , and are the four regions in counterclockwise order. Since , either is on side or and are on opposite sides of line . In either case and are on opposite sides of line . Also, since ray does not meet and ray does not meet , it follows that is entirely in the interior of . Hence and are on opposite sides of . Therefore is convex.
Second argument: Since and , cannot be inside or on the sides of triangle . Since , we have and hence . Hence cannot be inside or on the sides of triangle . Symmetrically, cannot be inside or on the sides of triangle . Finally, since and , we have Therefore cannot be inside or on the sides of triangle . Since this covers all four cases, is convex.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.