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 | + | If <math>r-t=s-t=0</math>, then <math>P</math> must be similar to <math>P'</math> and the conclusion is obvious. |
− | 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. | + | 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 (for the previous paragraph), WLOG assume the <math>(A'C'E')</math> is the unit circle, and compute <math>b'=a'+c'-\frac{a'c'}{e'}</math>, etc. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | We work in the complex plane to give (essentially) a complete characterization when we remove the condition that opposite sides are not parallel. | ||
+ | |||
+ | WLOG assume <math>a,b,c</math> are on the unit circle. It suffices to show that <math>a,b,c</math> uniquely determine <math>d,e,f</math>, since we know that if we let <math>E</math> be the reflection of <math>B</math> over <math>AC</math>, <math>D</math> be the reflection of <math>A</math> over <math>CE</math>, and <math>F</math> be the reflection of <math>C</math> over <math>AE</math>, then <math>ABCDEF</math> satisfies the problem conditions. (*) | ||
+ | |||
+ | It's easy to see with the given conditions that | ||
+ | <cmath>\begin{align*} | ||
+ | (a-b)(c-d)(e-f) &= (b-c)(d-e)(f-a) \Longleftrightarrow f=\frac{(a-b)(c-d)e+(c-b)(e-d)a}{(a-b)(c-d)+(c-b)(e-d)} \ | ||
+ | \frac{(e-a)(c-b)}{(a-b)(c-d)+(c-b)(e-d)} = \frac{f-e}{d-e} &= \left(\frac{c-b}{a-b}\right)^2 \overline{\left(\frac{a-b}{c-b}\right)} = \frac{c-b}{a-b}\cdot\frac{c}{a} \Longleftrightarrow d=\frac{c[(a-b)c+(c-b)e]+a(a-e)(a-b)}{c[(a-b)+(c-b)]} \ | ||
+ | \frac{(a-b)(c-d)+(c-b)(e-d)}{(a-e)(c-d)} = \frac{b-a}{f-a} &= \left(\frac{e-d}{c-d}\right)^2 \overline{\left(\frac{c-d}{e-d}\right)}. | ||
+ | \end{align*}</cmath> | ||
+ | Note that | ||
+ | <cmath>\frac{e-d}{c-d}=\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)},</cmath> | ||
+ | so plugging into the third equation we have | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{a(a-b)(2b-a-c)}{c(c-e)(c-b)-a(a-e)(a-b)} | ||
+ | &=\frac{(a-b)+(c-b)\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}}{(a-e)}\ | ||
+ | &=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\overline{\left(\frac{c(c-e)(c-b)-a(a-e)(a-b)}{(a-b)[c(e-c)+a(e-a)]}\right)}\ | ||
+ | &=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\frac{\frac{1}{c}\left(\overline{e}-\frac{1}{c}\right)\frac{b-c}{bc}-\frac{1}{a}\left(\overline{e}-\frac{1}{a}\right)\frac{b-a}{ba}}{\frac{b-a}{ab}\left(\frac{1}{c}\left(\frac{1}{c}-\overline{e}\right)+\frac{1}{a}\left(\frac{1}{a}-\overline{e}\right)\right)}\ | ||
+ | &=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\frac{c^3(a\overline{e}-1)(a-b)-a^3(c\overline{e}-1)(c-b)}{c(a-b)[a^2(c\overline{e}-1)+c^2(a\overline{e}-1)]}\ | ||
+ | &=\frac{(a-b)[c(e-c)+a(e-a)]^2}{[c(c-e)(c-b)-a(a-e)(a-b)]^2}\frac{c^3(a\overline{e}-1)(a-b)-a^3(c\overline{e}-1)(c-b)}{c[a^2(c\overline{e}-1)+c^2(a\overline{e}-1)]}. | ||
+ | \end{align*}</cmath> | ||
+ | Simplifying, this becomes | ||
+ | <cmath>\begin{align*} | ||
+ | &ac(2b-a-c)[c(c-e)(c-b)-a(a-e)(a-b)][a^2(c\overline{e}-1)+c^2(a\overline{e}-1)]\ | ||
+ | &=[c(e-c)+a(e-a)]^2[c^3(a\overline{e}-1)(a-b)-a^3(c\overline{e}-1)(c-b)]. | ||
+ | \end{align*}</cmath> | ||
+ | Of course, we can also "conjugate" this equation -- a nice way to do this is to note that if | ||
+ | <cmath>x=\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)},</cmath> | ||
+ | then | ||
+ | <cmath>\frac{a(2b-a-c)}{c(e-c)+a(e-a)}=\frac{x}{\overline{x}}=\overline{\left(\frac{c(e-c)+a(e-a)}{a(2b-a-c)}\right)}=\frac{\frac{1}{c}\left(\overline{e}-\frac{1}{c}\right)+\frac{1}{a}\left(\overline{e}-\frac{1}{a}\right)}{\frac{1}{a}\left(\frac{2}{b}-\frac{1}{a}-\frac{1}{c}\right)},</cmath> | ||
+ | whence | ||
+ | <cmath>ac(2b-a-c)[2ac-b(a+c)]=b[c(e-c)+a(e-a)][a^2(c\overline{e}-1)+c^2(a\overline{e}-1)].</cmath> | ||
+ | If <math>a+c\ne 0</math>, then eliminating <math>\overline{e}</math>, we get | ||
+ | <cmath>e\in\left\{a+c-\frac{ac}{b},a+\frac{2c(c-b)}{a+c},c+\frac{2a(a-b)}{a+c}\right\}.</cmath> | ||
+ | The first case corresponds to (*) (since <math>a,b,c,e</math> uniquely determine <math>d</math> and <math>f</math>), the second corresponds to <math>AB\parallel DE</math> (or equivalently, since <math>AB=DE</math>, <math>a-b=e-d</math>), and by symmetry, the third corresponds to <math>CB\parallel FE</math>. | ||
+ | |||
+ | Otherwise, if <math>c=-a</math>, then we easily find <math>b^2e=a^4\overline{e}</math> from the first of the two equations in <math>e,\overline{e}</math> (we actually don't need this, but it tells us that the locus of working <math>e</math> is a line through the origin). It's easy to compute <math>d=e+\frac{a(a-b)}{b}</math> and <math>f=e+\frac{a(a+b)}{b}</math>, so <math>a-c=2a=f-d\implies c-d=a-f\implies CD\parallel AF</math>, and we're done. | ||
+ | |||
+ | '''Comment.''' It appears that taking <math>(ABC)</math> the unit circle is nicer than, say <math>e=0</math> or <math>(ACE)</math> the unit circle (which may not even be reasonably tractable). | ||
==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 15:35, 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.
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 (for the previous paragraph), WLOG assume the is the unit circle, and compute , etc.
Solution 3
We work in the complex plane to give (essentially) a complete characterization when we remove the condition that opposite sides are not parallel.
WLOG assume are on the unit circle. It suffices to show that uniquely determine , since we know that if we let be the reflection of over , be the reflection of over , and be the reflection of over , then satisfies the problem conditions. (*)
It's easy to see with the given conditions that Note that so plugging into the third equation we have Simplifying, this becomes Of course, we can also "conjugate" this equation -- a nice way to do this is to note that if then whence If , then eliminating , we get The first case corresponds to (*) (since uniquely determine and ), the second corresponds to (or equivalently, since , ), and by symmetry, the third corresponds to .
Otherwise, if , then we easily find from the first of the two equations in (we actually don't need this, but it tells us that the locus of working is a line through the origin). It's easy to compute and , so , and we're done.
Comment. It appears that taking the unit circle is nicer than, say or the unit circle (which may not even be reasonably tractable).
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 |