Difference between revisions of "2001 USAMO Problems/Problem 4"
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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. | 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. | ||
− | ==Solution 4== | + | ===Solution 4=== |
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 | 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 | ||
<cmath>(-a + b + c)^2 \ge 0,</cmath> | <cmath>(-a + b + c)^2 \ge 0,</cmath> | ||
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<cmath>2a^2 - 2ab - 2ac + 2bc > a^2 + b^2 + c^2 - 2ab - 2ac + 2bc \ge 0</cmath> | <cmath>2a^2 - 2ab - 2ac + 2bc > a^2 + b^2 + c^2 - 2ab - 2ac + 2bc \ge 0</cmath> | ||
and therefore our desired conclusion. | and therefore our desired conclusion. | ||
+ | |||
+ | ===Solution 5=== | ||
+ | Let <math>M, N</math> be midpoints of <math>AP</math> and <math>BC</math>, respectively. For the points <math>A, B, P, C</math>; let's apply Euler's quadrilateral formula, | ||
+ | <cmath> AB^2 + BP^2 + PC^2 + CA^2 = AP^2 + BC^2 + 4MN^2 \geq AP^2 + BC^2 .</cmath> | ||
+ | Given that <math>AP^2 > BP^2 + PC^2</math>. Thus, | ||
+ | <cmath> AB^2 + AC^2 > BC^2 .</cmath> | ||
+ | and we get <math>\angle BAC</math> is acute. | ||
+ | |||
+ | (Lokman GÖKÇE) | ||
+ | |||
+ | ===Solution 6=== | ||
+ | Without loss of generality, assume that in a Cartesian coordinate system, <math>A</math> is at the point <math>(0, 0)</math> and <math>C</math> is at the point <math>(1,0)</math>. Let <math>B</math> be at the point <math>(b_x,b_y)</math> and <math>P</math> be at the point <math>(p_x,p_y)</math>. Without loss of generality, also assume that <math>b_y>0</math>. | ||
+ | |||
+ | Now, assume for contradiction that <math>\angle BAC</math> is not acute. Since <math>PA</math>, <math>PB</math>, and <math>PC</math> are the sides of an obtuse triangle, with <math>PA</math> the longest side, it follows that <math>PA^2>PB^2+PC^2</math>, implying that <math>p_x^2+p_y^2>(p_x-b_x)^2+(p_y-b_y)^2+(p_x-1)^2+p_y^2</math>. This inequality simplifies to <math>b_x^2-2p_x b_x+b_y^2-2p_y b_y+p_x^2-2p_x+1+p_y^2<0</math>. Note that since <math>p_x^2-2p_x+1</math> and <math>b_y^2-2p_y b_y+p_y^2</math> are both perfect squares, all terms of this inequality except for <math>-2p_x b_x</math> are already guaranteed to be nonnegative. | ||
+ | |||
+ | If <math>p_x<0</math>, then <math>P</math> would be closer to <math>A</math> than to <math>C</math>, but since <math>PA^2=PB^2+PC^2</math>, this is not possible. Therefore, <math>p_x \geq 0</math>. Since <math>\angle BAC</math> not being acute implies that <math>b_x \leq 0</math>, it follows that <math>-2p_x b_x \geq 0</math>. But now since all terms of <math>b_x^2-2p_x b_x+b_y^2-2p_y b_y+p_x^2-2p_x+1+p_y^2<0</math> are guaranteed to be nonnegative, this entire expression cannot be negative, leading to a contradiction. Therefore, <math>\angle BAC</math> is acute. | ||
== See also == | == See also == |
Latest revision as of 21:24, 18 June 2022
Contents
[hide]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
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.
Solution 4
Let be the origin in vector space, and let denote the position vectors of respectively. Then the obtuse triangle condition, , becomes 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 is acute, it suffices to show that , or . But this follows from the observation that which leads to and therefore our desired conclusion.
Solution 5
Let be midpoints of and , respectively. For the points ; let's apply Euler's quadrilateral formula, Given that . Thus, and we get is acute.
(Lokman GÖKÇE)
Solution 6
Without loss of generality, assume that in a Cartesian coordinate system, is at the point and is at the point . Let be at the point and be at the point . Without loss of generality, also assume that .
Now, assume for contradiction that is not acute. Since , , and are the sides of an obtuse triangle, with the longest side, it follows that , implying that . This inequality simplifies to . Note that since and are both perfect squares, all terms of this inequality except for are already guaranteed to be nonnegative.
If , then would be closer to than to , but since , this is not possible. Therefore, . Since not being acute implies that , it follows that . But now since all terms of are guaranteed to be nonnegative, this entire expression cannot be negative, leading to a contradiction. Therefore, is acute.
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.