Difference between revisions of "2001 IMO Shortlist Problems/G1"
(New page: == Problem == Let <math>A_1</math> be the center of the square inscribed in acute triangle <math>ABC</math> with two vertices of the square on side <math>BC</math>. Thus one of the two re...) |
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== Solution == | == Solution == | ||
− | {{ | + | Let <math>BC=a</math>, <math>CA=b</math>, <math>AB=c</math>, <math>\angle BAC=\alpha</math>, <math>\angle CBA=\beta</math>, and <math>\angle ACB=\gamma</math>. Let <math>A_2</math> be the point on the other side of <math>BC</math> than <math>A</math> such that <math>BA_2C</math> is an isosceles right triangle. Define <math>B_2</math> and <math>C_2</math> similarly. Let <math>E</math> and <math>F</math> be the points on <math>AB</math> and <math>CA</math> that are the vertices of the square centered at <math>A_1</math>. We then have that <math>EA_1F</math> is also an isosceles right triangle. It's clear that <math>EF</math> is parallel to <math>BC</math>, so <math>\triangle AEF\sim \triangle ABC</math> and <math>\triangle A_1EF\sim \triangle A_2BC</math>. The ratio of similarity of both relations is <math>\frac{EF}{BC}</math>, which implies that quadrilaterals <math>AEA_1F</math> and <math>ABA_2C</math> are similar. Therefore <math>\angle BAA_2=\angle EAA_1</math> and <math>\angle CAA_2=EAA_1</math>. It then follows that <math>A</math>, <math>A_1</math>, and <math>A_2</math> are collinear. Similarly <math>B</math>, <math>B_1</math>, and <math>B_2</math> are collinear, as are <math>C</math>, <math>C_1</math>, and <math>C_2</math>. It therefore suffices to show that <math>AA_2</math>, <math>BB_2</math>, and <math>CC_2</math> are concurrent. |
+ | |||
+ | Let <math>AC_2=s</math> and <math>CC_2=d</math>, for positive reals <math>s</math> and <math>d</math>. Also let <math>\angle ACC_2=\theta</math> and <math>\angle BCC_2=\phi</math>. Note that <math>\angle C_2AC=\angle C_2AB+\angle BAC=45^{\circ}+\alpha</math>, and likewise <math>\angle C_2BC = 45^{\circ}+\beta</math>. It then follows from the Law of Sines on triangles <math>ACC_2</math> and <math>BCC_2</math> that | ||
+ | |||
+ | <cmath>\frac{d}{\sin{(\alpha+45^{\circ})}}=\frac{s}{\sin{\theta}}</cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | <cmath>\frac{d}{\sin{(\beta+45^{\circ})}}=\frac{s}{\sin{\phi}}</cmath> | ||
+ | |||
+ | Solving for <math>\sin{\theta}</math> and <math>\sin{\phi}</math> gives that | ||
+ | |||
+ | <cmath>\sin{\theta}=\frac{s\sin{(\alpha+45^{\circ})}}{d}</cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | <cmath>\sin{\phi}=\frac{s\sin{(\beta+45^{\circ})}}{d}</cmath> | ||
+ | |||
+ | Therefore | ||
+ | |||
+ | <cmath>\frac{\sin{\theta}}{\sin{\phi}}=\frac{\sin\angle ACC_2}{\sin\angle BCC_2}=\frac{\sin{(\alpha+45^{\circ})}}{\sin{(\beta+45^{\circ})}}</cmath> | ||
+ | |||
+ | Similar lines of reasoning show that | ||
+ | |||
+ | <cmath>\frac{\sin\angle BAA_2}{\sin\angle CAA_2}=\frac{\sin{(\beta+45^{\circ})}}{\sin{(\gamma+45^{\circ})}}</cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | <cmath>\frac{\sin\angle CBB_2}{\sin\angle ABB_2}=\frac{\sin{(\gamma+45^{\circ})}}{\sin{(\alpha+45^{\circ})}}</cmath> | ||
+ | |||
+ | An application of the trigonometric version of Ceva's Theorem shows that <math>AA_2</math>, <math>BB_2</math>, and <math>CC_2</math> are concurrent, which shows that <math>AA_1</math>, <math>BB_1</math>, and <math>CC_1</math> are concurrent. | ||
== Resources == | == Resources == |
Latest revision as of 15:07, 12 December 2011
Problem
Let be the center of the square inscribed in acute triangle
with two vertices of the square on side
. Thus one of the two remaining vertices of the square is on side
and the other is on
. Points
are defined in a similar way for inscribed squares with two vertices on sides
and
, respectively. Prove that lines
are concurrent.
Solution
Let ,
,
,
,
, and
. Let
be the point on the other side of
than
such that
is an isosceles right triangle. Define
and
similarly. Let
and
be the points on
and
that are the vertices of the square centered at
. We then have that
is also an isosceles right triangle. It's clear that
is parallel to
, so
and
. The ratio of similarity of both relations is
, which implies that quadrilaterals
and
are similar. Therefore
and
. It then follows that
,
, and
are collinear. Similarly
,
, and
are collinear, as are
,
, and
. It therefore suffices to show that
,
, and
are concurrent.
Let and
, for positive reals
and
. Also let
and
. Note that
, and likewise
. It then follows from the Law of Sines on triangles
and
that
and
Solving for and
gives that
and
Therefore
Similar lines of reasoning show that
and
An application of the trigonometric version of Ceva's Theorem shows that ,
, and
are concurrent, which shows that
,
, and
are concurrent.