Difference between revisions of "2001 USAMO Problems/Problem 4"
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<cmath>\angle ABP + \angle ACP > \angle BAC\geq 90^\circ\geq\angle ABC + \angle ACB.</cmath> | <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. | 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. | ||
== See also == | == See also == |
Revision as of 13:58, 16 February 2015
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
(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.
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.
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.