Difference between revisions of "2011 USAMO Problems/Problem 3"
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&=(r-t)(a'-b'+d'-e')+(s-t)(b'-c'+e'-f'). | &=(r-t)(a'-b'+d'-e')+(s-t)(b'-c'+e'-f'). | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | If <math>r-t=s-t=0</math>, then <math>P</math> must be similar to <math>P'</math> and the conclusion is obvious. Otherwise, since <math>a'-b'+d'-e'\ne0</math> and <math>b'-c'+e'-f'\ne0</math>, we must have <math>r-t\ne0</math> and <math>s-t\ne0</math>. Construct parallelograms <math>A'XE'D'</math> and <math>C'YD'E'</math>; if <math>U</math> is the reflection of <math>A'</math> over <math>B'X</math> and <math>V</math> is the reflection of <math>C'</math> over <math>D'Y</math>, then by simple angle chasing we can show that <math>\angle{UA'C'}=180^\circ-\angle{A'E'C'}</math> and <math>\angle{VC'A'}=180^\circ-\angle{C'E'A'}</math>. But <math>(r-t)(s-t)\ne0</math> means <math>u-a'=b'-a'+e'-d'</math> and <math>v-c'=b'-c'+e'-f'</math> must be linearly dependent (note that <math>A'B'=A'X</math> and <math>C'D'=C'V</math>), so we must have <math>\angle{UA'C'}+\angle{VC'A'}=180^\circ\implies \angle{A'E'C'}=90^\circ</math>. But then C'D'\parallel A'F'<math>, which is impossible, so we're done. | + | If <math>r-t=s-t=0</math>, then <math>P</math> must be similar to <math>P'</math> and the conclusion is obvious. Otherwise, since <math>a'-b'+d'-e'\ne0</math> and <math>b'-c'+e'-f'\ne0</math>, we must have <math>r-t\ne0</math> and <math>s-t\ne0</math>. Construct parallelograms <math>A'XE'D'</math> and <math>C'YD'E'</math>; if <math>U</math> is the reflection of <math>A'</math> over <math>B'X</math> and <math>V</math> is the reflection of <math>C'</math> over <math>D'Y</math>, then by simple angle chasing we can show that <math>\angle{UA'C'}=180^\circ-\angle{A'E'C'}</math> and <math>\angle{VC'A'}=180^\circ-\angle{C'E'A'}</math>. But <math>(r-t)(s-t)\ne0</math> means <math>u-a'=b'-a'+e'-d'</math> and <math>v-c'=b'-c'+e'-f'</math> must be linearly dependent (note that <math>A'B'=A'X</math> and <math>C'D'=C'V</math>), so we must have <math>\angle{UA'C'}+\angle{VC'A'}=180^\circ\implies \angle{A'E'C'}=90^\circ</math>. But then <math>C'D'\parallel A'F'</math>, which is impossible, so we're done. |
− | Alternatively, | + | Alternatively, WLOG assume the <math>(A'C'E')</math> is the unit circle, and compute <math>b'=a'+c'-\frac{a'c'}{e'}</math>, etc. |
==See Also== | ==See Also== | ||
{{USAMO newbox|year=2011|num-b=2|num-a=4}} | {{USAMO newbox|year=2011|num-b=2|num-a=4}} |
Revision as of 15:25, 29 October 2012
In hexagon , which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy , , and . Furthermore , , and . Prove that diagonals , , and are concurrent.
Contents
[hide]Solutions
Solution 1
Let , , and , , , , intersect at , intersect at , and intersect at . Define the vectors: Clearly, .
Note that . By sliding the vectors and to the vectors and respectively, then . As is isosceles with , the base angles are both . Thus, . Similarly, and .
Next we will find the angles between , , and . As , the angle between the vectors and is . Similarly, the angle between and is , and the angle between and is . Thus, the angle between and is , or just in the other direction if we take it modulo . Similarly, the angle between and is , and the angle between and is .
And since , we can arrange the three vectors to form a triangle, so the triangle with sides of lengths , , and has opposite angles of , , and , respectively. So by the law of sines: and the triangle with sides of length , , and has corrosponding angles of , , and . But then triangles , , and . So , , and , and , , and are the reflections of the vertices of triangle about the sides. So , , and concur at the orthocenter of triangle .
Solution 2
We work in the complex plane, where lowercase letters denote point affixes. Let denote hexagon . Since , the condition is equivalent to .
Construct a "phantom hexagon" as follows: let be a triangle with , , and (this is possible since by the angle conditions), and reflect over its sides to get points , respectively. By rotation and reflection if necessary, we assume and have the same orientation (clockwise or counterclockwise), i.e. . It's easy to verify that for and opposite sides of have equal lengths. As the corresponding sides of and must then be parallel, there exist positive reals such that , , and . But then , etc., so If , then must be similar to and the conclusion is obvious. Otherwise, since and , we must have and . Construct parallelograms and ; if is the reflection of over and is the reflection of over , then by simple angle chasing we can show that and . But means and must be linearly dependent (note that and ), so we must have . But then , which is impossible, so we're done.
Alternatively, WLOG assume the is the unit circle, and compute , etc.
See Also
2011 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |