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Difference between revisions of "1972 USAMO Problems/Problem 5"
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<asy>
 
<asy>
−
size(80);
+
size(120);
−
defaultpen(fontsize(7));
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defaultpen(fontsize(10));
−
pair A=(0,7), B=(5,4), C=(3,0), D=(-3,0), E=(-5,4), P;
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pair A=dir(90), B=dir(90-72), C=dir(90-2*72), D=dir(90-3*72), E=dir(90-4*72);
−
P=extension(B,D,C,E);
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draw(A--B--C--D--E--cycle);
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draw(D--E--A--B--C--D--B--E--C);
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draw(A--C--E--B--D--cycle);
−
label("A",A,(0,1));label("B",B,(1,0));label("C",C,(1,-1));label("D",D,(-1,-1));label("E",E,(-1,0));label("P",P,(0,1));
+
label("A",A,A);label("B",B,B);label("C",C,C);label("D",D,D);label("E",E,E);
 
</asy>
 
</asy>
  
 
==Solution==
 
==Solution==
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{{solution}}
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'''Lemma:''' Convex pentagon <math>A_0A_1A_2A_3A_4</math> has the property that <math>[A_0A_1A_2] = [A_1A_2A_3] = [A_2A_3A_4] = [A_3A_4A_0] = [A_4A_0A_1]</math> if and only if <math>\overline{A_{n - 1}A_{n + 1}}\parallel\overline{A_{n - 2}A_{n + 2}}</math> for <math>n = 0, 1, 2, 3, 4</math> (indices taken mod 5).
 +
 
 +
'''Proof:''' For the "only if" direction, since <math>[A_0A_1A_2] = [A_1A_2A_3]</math>, <math>A_0</math> and <math>A_3</math> are equidistant from <math>\overline{A_1A_2}</math>, and since the pentagon is convex, <math>\overline{A_0A_3}\parallel\overline{A_1A_2}</math>. The other four pairs of parallel lines are established by a symmetrical argument, and the proof for the other direction is just this, but reversed.
 +
 
 +
<asy>
 +
size(120);
 +
defaultpen(fontsize(10));
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pathpen = black;
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pair A=MP("A",dir(90),dir(90)), B=MP("B",dir(90-72),dir(90-72)), C=MP("C",dir(90-2*72),dir(90-2*72)), D=MP("D",dir(90-3*72),dir(90-3*72)), E=MP("E",dir(90-4*72),dir(90-4*72));
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D(A--B--C--D--E--cycle);
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D(A--C--E--B--D--cycle);
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pair Ap = MP("A'",IP(B--D,C--E),dir(270)), Bp = MP("B'",IP(A--D,C--E),dir(270-72)), Cp = MP("C'",IP(A--D,B--E),dir(270-72*2)), Dp = MP("D'",IP(A--C,B--E),dir(270-72*3)), Ep = MP("E'",IP(A--C,B--D),dir(270-72*4));
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</asy>
 +
 
 +
Let <math>A'B'C'D'E'</math> be the inner pentagon, labeled so that <math>A</math> and <math>A'</math> are opposite each other, and let <math>a, b, c, d, e</math> be the side lengths of the inner pentagon labeled opposite their corresponding vertices. From the lemma, pentagon <math>A'B'C'D'E'</math> is similar to pentagon <math>ABCDE</math> with <math>AB = m(A'B')</math> and parallelogram <math>ABCB'</math> (and cyclic) has area 2. Supposing
 +
<cmath>\begin{align*}
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P &= [ABCDE] \\
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Q &= [A'B'C'D'E'] \\
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R &= \sum_{\text{cyc}}[AC'D'] \\
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S &= \sum_{\text{cyc}}[ABD'],
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\end{align*}</cmath>
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we have
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<cmath>\begin{align*}
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\sum_{\text{cyc}}[ABCB'] &= 5Q + 3R + 2S \\
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10 &= 3Q + R + 2P.
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\end{align*}</cmath>
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Since <math>BCDC'</math> and <math>CDED'</math> are parallelograms, <math>BD' = EC' = (m - 1)a</math>. Triangle <math>EB'C'</math> is similar to triangle <math>ECB</math>, so <math>me = BC = \left(\frac{2m - 1}{m - 1}\right)e</math>, and with the requirement that <math>m > 1</math>,
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<cmath>m = \frac{2m - 1}{m - 1}\Rightarrow m = \frac{3 + \sqrt{5}}{2}.</cmath>
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Now, we compute that <math>[AC'D'] = \frac{1}{2m - 1} = \sqrt{5} - 2</math>, and similar computation for the other four triangles gives <math>R = 5\sqrt{5} - 10</math>. From the aforementioned pentagon similarity, <math>Q = P/m^2 = \left(\frac{7 - 3\sqrt{5}}{2}\right)P</math>. Solving for <math>P</math>, we have
 +
<cmath>\begin{align*}
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10 &= 3\left(\frac{7 - 3\sqrt{5}}{2}\right)P + (5\sqrt{5} - 10) + 2P \\
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20 - 5\sqrt{5} &= \left(\frac{25 - 9\sqrt{5}}{2}\right)P \\
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P &= \frac{40 - 10\sqrt{5}}{25 - 9\sqrt{5}} \\
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&= \frac{5 + \sqrt{5}}{2}.
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\end{align*}</cmath>
 +
 
 +
To show that there are infinitely many pentagons with the given property, we start with triangle <math>A'B'C'</math> and construct the original pentagon using the required length ratios. This triangle has three continuous degrees of freedom, and the condition that the equal areas be specifically unity means that there are two continuous degrees of freedom in which the configuration can vary, which implies the desired result.
 +
 
 +
Alternatively, one can shear a regular pentagon, a transformation which preserves areas and the property of a pair of lines being parallel.
  
 
==See Also==
 
==See Also==
  
 
{{USAMO box|year=1972|num-b=4|after=Last Question}}
 
{{USAMO box|year=1972|num-b=4|after=Last Question}}
 +
{{MAA Notice}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Latest revision as of 08:29, 26 March 2023

Problem

A given convex pentagon $ABCDE$ has the property that the area of each of the five triangles $ABC$, $BCD$, $CDE$, $DEA$, and $EAB$ is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the above area property.

[asy] size(120); defaultpen(fontsize(10)); pair A=dir(90), B=dir(90-72), C=dir(90-2*72), D=dir(90-3*72), E=dir(90-4*72); draw(A--B--C--D--E--cycle); draw(A--C--E--B--D--cycle); label("A",A,A);label("B",B,B);label("C",C,C);label("D",D,D);label("E",E,E); [/asy]

Solution

Lemma: Convex pentagon $A_0A_1A_2A_3A_4$ has the property that $[A_0A_1A_2] = [A_1A_2A_3] = [A_2A_3A_4] = [A_3A_4A_0] = [A_4A_0A_1]$ if and only if $\overline{A_{n - 1}A_{n + 1}}\parallel\overline{A_{n - 2}A_{n + 2}}$ for $n = 0, 1, 2, 3, 4$ (indices taken mod 5).

Proof: For the "only if" direction, since $[A_0A_1A_2] = [A_1A_2A_3]$, $A_0$ and $A_3$ are equidistant from $\overline{A_1A_2}$, and since the pentagon is convex, $\overline{A_0A_3}\parallel\overline{A_1A_2}$. The other four pairs of parallel lines are established by a symmetrical argument, and the proof for the other direction is just this, but reversed.

[asy] size(120); defaultpen(fontsize(10)); pathpen = black; pair A=MP("A",dir(90),dir(90)), B=MP("B",dir(90-72),dir(90-72)), C=MP("C",dir(90-2*72),dir(90-2*72)), D=MP("D",dir(90-3*72),dir(90-3*72)), E=MP("E",dir(90-4*72),dir(90-4*72)); D(A--B--C--D--E--cycle); D(A--C--E--B--D--cycle); pair Ap = MP("A'",IP(B--D,C--E),dir(270)), Bp = MP("B'",IP(A--D,C--E),dir(270-72)), Cp = MP("C'",IP(A--D,B--E),dir(270-72*2)), Dp = MP("D'",IP(A--C,B--E),dir(270-72*3)), Ep = MP("E'",IP(A--C,B--D),dir(270-72*4)); [/asy]

Let $A'B'C'D'E'$ be the inner pentagon, labeled so that $A$ and $A'$ are opposite each other, and let $a, b, c, d, e$ be the side lengths of the inner pentagon labeled opposite their corresponding vertices. From the lemma, pentagon $A'B'C'D'E'$ is similar to pentagon $ABCDE$ with $AB = m(A'B')$ and parallelogram $ABCB'$ (and cyclic) has area 2. Supposing \begin{align*} P &= [ABCDE] \\ Q &= [A'B'C'D'E'] \\ R &= \sum_{\text{cyc}}[AC'D'] \\ S &= \sum_{\text{cyc}}[ABD'], \end{align*} we have \begin{align*} \sum_{\text{cyc}}[ABCB'] &= 5Q + 3R + 2S \\ 10 &= 3Q + R + 2P. \end{align*} Since $BCDC'$ and $CDED'$ are parallelograms, $BD' = EC' = (m - 1)a$. Triangle $EB'C'$ is similar to triangle $ECB$, so $me = BC = \left(\frac{2m - 1}{m - 1}\right)e$, and with the requirement that $m > 1$, \[m = \frac{2m - 1}{m - 1}\Rightarrow m = \frac{3 + \sqrt{5}}{2}.\] Now, we compute that $[AC'D'] = \frac{1}{2m - 1} = \sqrt{5} - 2$, and similar computation for the other four triangles gives $R = 5\sqrt{5} - 10$. From the aforementioned pentagon similarity, $Q = P/m^2 = \left(\frac{7 - 3\sqrt{5}}{2}\right)P$. Solving for $P$, we have \begin{align*} 10 &= 3\left(\frac{7 - 3\sqrt{5}}{2}\right)P + (5\sqrt{5} - 10) + 2P \\ 20 - 5\sqrt{5} &= \left(\frac{25 - 9\sqrt{5}}{2}\right)P \\ P &= \frac{40 - 10\sqrt{5}}{25 - 9\sqrt{5}} \\ &= \frac{5 + \sqrt{5}}{2}. \end{align*}

To show that there are infinitely many pentagons with the given property, we start with triangle $A'B'C'$ and construct the original pentagon using the required length ratios. This triangle has three continuous degrees of freedom, and the condition that the equal areas be specifically unity means that there are two continuous degrees of freedom in which the configuration can vary, which implies the desired result.

Alternatively, one can shear a regular pentagon, a transformation which preserves areas and the property of a pair of lines being parallel.

See Also

1972 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Last Question
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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