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 | ||
Line 11: | Line 11: | ||
Then, we wish to show | Then, we wish to show | ||
− | < | + | <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> |
− | (p-1)^2 + q^2 + (p-x)^2 + (q-y)^2 + 1 + x^2 + y^2 | + | <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> |
− | 2p^2 + 2q^2 + 2x^2 + 2y^2 - 2p - 2px - 2qy + 2 | + | <cmath>p^2 + q^2 + x^2 + y^2 + 2x - 2p - 2px - 2qy + 1 \geq 0 </cmath> |
− | p^2 + q^2 + x^2 + y^2 + 2x - 2p - 2px - 2qy + 1 | + | <cmath>(x-p)^2 + (q-y)^2 + 2(x-p) + 1 \geq 0 </cmath> |
− | (x-p)^2 + (q-y)^2 + 2(x-p) + 1 | + | <cmath>(x-p+1)^2 + (q-y)^2 \geq 0,</cmath> |
− | (x-p+1)^2 + (q-y)^2 | ||
− | |||
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. | ||
+ | |||
+ | ===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 | ||
+ | <cmath>(-a + b + c)^2 \ge 0,</cmath> | ||
+ | which leads to | ||
+ | <cmath>2a^2 - 2ab - 2ac + 2bc > a^2 + b^2 + c^2 - 2ab - 2ac + 2bc \ge 0</cmath> | ||
+ | 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 == | ||
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[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
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.