Difference between revisions of "2011 USAMO Problems/Problem 3"
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In hexagon <math>ABCDEF</math>, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy <math>\angle A = 3\angle D</math>, <math>\angle C = 3\angle F</math>, and <math>\angle E = 3\angle B</math>. Furthermore <math>AB=DE</math>, <math>BC=EF</math>, and <math>CD=FA</math>. Prove that diagonals <math>\overline{AD}</math>, <math>\overline{BE}</math>, and <math>\overline{CF}</math> are concurrent. | In hexagon <math>ABCDEF</math>, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy <math>\angle A = 3\angle D</math>, <math>\angle C = 3\angle F</math>, and <math>\angle E = 3\angle B</math>. Furthermore <math>AB=DE</math>, <math>BC=EF</math>, and <math>CD=FA</math>. Prove that diagonals <math>\overline{AD}</math>, <math>\overline{BE}</math>, and <math>\overline{CF}</math> are concurrent. | ||
− | ==Solution== | + | ==Solutions== |
+ | ===Solution 1=== | ||
Let <math>\angle A = \alpha</math>, <math>\angle C = \gamma</math>, and <math>\angle E = \beta</math>, <math>AB=DE=p</math>, <math>BC=EF=q</math>, <math>CD=FA=r</math>, <math>AB</math> intersect <math>DE</math> at <math>X</math>, <math>BC</math> intersect <math>EF</math> at <math>Y</math>, and <math>CD</math> intersect <math>FA</math> at <math>Z</math>. Define the vectors: <cmath>\vec{u} = \vec{AB} + \vec{DE}</cmath> <cmath>\vec{v} = \vec{BC} + \vec{EF}</cmath> <cmath>\vec{w} = \vec{CD} + \vec{FA}</cmath> Clearly, <math>\vec{u}+\vec{v}+\vec{w}=\vec{0}</math>. | Let <math>\angle A = \alpha</math>, <math>\angle C = \gamma</math>, and <math>\angle E = \beta</math>, <math>AB=DE=p</math>, <math>BC=EF=q</math>, <math>CD=FA=r</math>, <math>AB</math> intersect <math>DE</math> at <math>X</math>, <math>BC</math> intersect <math>EF</math> at <math>Y</math>, and <math>CD</math> intersect <math>FA</math> at <math>Z</math>. Define the vectors: <cmath>\vec{u} = \vec{AB} + \vec{DE}</cmath> <cmath>\vec{v} = \vec{BC} + \vec{EF}</cmath> <cmath>\vec{w} = \vec{CD} + \vec{FA}</cmath> Clearly, <math>\vec{u}+\vec{v}+\vec{w}=\vec{0}</math>. | ||
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And since <math>\vec{u}+\vec{v}+\vec{w}=\vec{0}</math>, we can arrange the three vectors to form a triangle, so the triangle with sides of lengths <math>2p \cos \gamma</math>, <math>2q \cos \alpha</math>, and <math>2r \cos \beta</math> has opposite angles of <math>180^\circ - 2\gamma</math>, <math>180^\circ - 2\alpha</math>, and <math>180^\circ - 2\beta</math>, respectively. So by the law of sines: <cmath> \frac{2p \cos \gamma}{\sin 2\gamma} = \frac{2q \cos \alpha}{\sin 2\alpha} = \frac{2r \cos \beta}{\sin 2\beta} </cmath> <cmath> \frac{p}{\sin \gamma} = \frac{q}{\sin \alpha} = \frac{r}{\sin \beta}, </cmath> and the triangle with sides of length <math>p</math>, <math>q</math>, and <math>r</math> has corrosponding angles of <math>\gamma</math>, <math>\alpha</math>, and <math>\beta</math>. But then triangles <math>FAB</math>, <math>CDB</math>, and <math>FDE</math>. So <math>FD=p</math>, <math>BF=q</math>, and <math>BD=r</math>, and <math>A</math>, <math>C</math>, and <math>E</math> are the reflections of the vertices of triangle <math>BDF</math> about the sides. So <math>AD</math>, <math>BE</math>, and <math>CF</math> concur at the orthocenter of triangle <math>BDF</math>. | And since <math>\vec{u}+\vec{v}+\vec{w}=\vec{0}</math>, we can arrange the three vectors to form a triangle, so the triangle with sides of lengths <math>2p \cos \gamma</math>, <math>2q \cos \alpha</math>, and <math>2r \cos \beta</math> has opposite angles of <math>180^\circ - 2\gamma</math>, <math>180^\circ - 2\alpha</math>, and <math>180^\circ - 2\beta</math>, respectively. So by the law of sines: <cmath> \frac{2p \cos \gamma}{\sin 2\gamma} = \frac{2q \cos \alpha}{\sin 2\alpha} = \frac{2r \cos \beta}{\sin 2\beta} </cmath> <cmath> \frac{p}{\sin \gamma} = \frac{q}{\sin \alpha} = \frac{r}{\sin \beta}, </cmath> and the triangle with sides of length <math>p</math>, <math>q</math>, and <math>r</math> has corrosponding angles of <math>\gamma</math>, <math>\alpha</math>, and <math>\beta</math>. But then triangles <math>FAB</math>, <math>CDB</math>, and <math>FDE</math>. So <math>FD=p</math>, <math>BF=q</math>, and <math>BD=r</math>, and <math>A</math>, <math>C</math>, and <math>E</math> are the reflections of the vertices of triangle <math>BDF</math> about the sides. So <math>AD</math>, <math>BE</math>, and <math>CF</math> concur at the orthocenter of triangle <math>BDF</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | We work in the complex plane, where lowercase letters denote point affixes. Let <math>P</math> denote hexagon <math>ABCDEF</math>. Since <math>AB=DE</math>, the condition <math>AB\not\parallel DE</math> is equivalent to <math>a-b+d-e\ne 0</math>. | ||
+ | |||
+ | Construct a "phantom hexagon" <math>P'=A'B'C'D'E'F'</math> as follows: let <math>A'C'E'</math> be a triangle with <math>\angle{A'C'E'}=\angle{F}</math>, <math>\angle{C'E'A'}=\angle{B}</math>, and <math>\angle{E'A'C'}=\angle{F}</math> (this is possible since <math>\angle{B}+\angle{D}+\angle{F}=180^\circ</math> by the angle conditions), and reflect <math>A',C',E'</math> over its sides to get points <math>D',F',B'</math>, respectively. By rotation and reflection if necessary, we assume <math>A'B'\parallel AB</math> and <math>P',P</math> have the same orientation (clockwise or counterclockwise), i.e. <math>\frac{b-a}{b'-a'}\in\mathbb{R}^+</math>. It's easy to verify that <math>\angle{X'}=\angle{X}</math> for <math>X\in\{A,B,C,D,E,F\}</math> and opposite sides of <math>P'</math> have equal lengths. As the corresponding sides of <math>P</math> and <math>P'</math> must then be parallel, there exist positive reals <math>r,s,t</math> such that <math>r=\frac{a-b}{a'-b'}=\frac{d-e}{d'-e'}</math>, <math>s=\frac{b-c}{b'-c'}=\frac{e-f}{e'-f'}</math>, and <math>t=\frac{c-d}{c'-d'}=\frac{f-a}{f'-a'}</math>. But then <math>0\ne a-b+d-e=r(a'-b'+d'-e')</math>, etc., so | ||
+ | <cmath>\begin{align*} | ||
+ | 0 | ||
+ | &=(a-b+d-e)+(b-c+e-f)+(c-d+f-a) \ | ||
+ | &=r(a'-b'+d'-e')+s(b'-c'+e'-f')+t(c'-d'+f'-a') \ | ||
+ | &=(r-t)(a'-b'+d'-e')+(s-t)(b'-c'+e'-f'). | ||
+ | \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. | ||
+ | |||
+ | Alternatively, use the projection formula with unit circle </math>(A'C'E')$. | ||
+ | |||
==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 14:20, 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.
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[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 C'D'\parallel A'F'$, which is impossible, so we're done.
Alternatively, use the projection formula with unit circle$ (Error compiling LaTeX. Unknown error_msg)(A'C'E')$.
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 |