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2003 AIME I Problems/Problem 12: Difference between revisions

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Notice that <math>AB = CD</math>, and <math>BD = DB</math>, and <math>\angle{DAB} \cong \angle{BCD}</math>, so we have side-side-angle matching on triangles <math>ABD</math> and <math>CDB</math>.  Since the problem does not allow <math>\triangle{ABD} \cong \triangle{CDB}</math>, we know that <math>\angle{ADB}</math> is not a right angle, and there is a unique other triangle with the matching side-side-angle.
Notice that <math>AB = CD</math>, and <math>BD = DB</math>, and <math>\angle{DAB} \cong \angle{BCD}</math>, so we have side-side-angle matching on triangles <math>ABD</math> and <math>CDB</math>.  Since the problem does not allow <math>\triangle{ABD} \cong \triangle{CDB}</math>, we know that <math>\angle{ADB}</math> is not a right angle, and there is a unique other triangle with the matching side-side-angle.


Extend <math>AD</math> to <math>C'</math> so that <math>\triangle{ABC'}</math> is isosceles with <math>AB = C'B</math>. Then notice that <math>\triangle{DC'B}</math> has matching side-side-angle, and yet <math>\triangle{ADB} \not\cong \triangle{C'DB}</math> because <math>\angle{ADB}</math> is not right.  Therefore <math>\triangle{C'DB}</math> is the unique triangle mentioned above, so <math>\triangle{CDB}</math> is congruent, in some order of vertices, to <math>\triangle{C'DB}</math>. Since <math>\triangle{CDB} \cong \triangle{C'DB}</math> would imply <math>\triangle{CDB} = \triangle{C'DB}</math>, making quadrilateral <math>ABCD</math> degenerate, we must have <math>\triangle{CDB} \cong \triangle{C'BD}</math>.
<center><asy>
pair A,B,C,D,F;
A=(0,0);B=(5,4.04061);C=(6,0);D=(4,0);F=(5,0);
draw(B--A--C--B--D);
label("A",A,SW,red);label("E",F,S);label("B",B,NW,red);label("D",D,S,red);label("B",C,SE,blue);label("C",A,NW,blue);label("D",B,NE,blue);label("$\theta$",A,(4.5,1.5));
draw(anglemark(D,A,B,20));label("180",(A+B)/2,NW);
draw(B--F,dashed);draw(rightanglemark(B,F,A));draw(D--B--C,dashed);
add(pathticks(B--C,1,0.5,1,10));add(pathticks(B--D,1,0.5,1,10));
</asy></center>
Overlay the triangles <math>\triangle BDA</math> and <math>\triangle BDC</math> on each other as in the diagram above (where the red labels correspond to <math>\triangle BDA</math> and the blue labels correspond to <math>\triangle BDC</math>). Here we assume without loss of generality that <math>BC>AD</math>. Furthermore, let <math>\theta</math> be the angle <math>A</math> referenced in the problem; we need to find <math>\lfloor 1000\cos\theta\rfloor</math>.


Since the perimeter of <math>ABCD</math> is <math>640</math>, <math>AD + BC = 640 - 180 - 180 = 280</math>. Hence <math>280 = AD + BC = AD + DC'</math>. Drop the altitude of <math>\triangle{ABC'}</math> from <math>B</math> and call the foot <math>P</math>. Then right triangle trigonometry on <math>\triangle{APB}</math> shows that <math>\cos{A} = AP/AB = 140/180 = 7/9</math>, so <math>\lfloor 1000 \cos A \rfloor = \boxed{777}</math>.
Since the perimeter of <math>ABCD</math> is <math>640</math>, we have <math>AD+BC=640-360=280</math>. Thus let <math>BC=a</math> and <math>AD=280-a</math> for some positive real number <math>a</math>. But the sides that correspond to <math>\overline{BD}</math> above are congruent, so we can drop a perpendicular from the topmost point to a point <math>E</math>, where the base of the isosceles triangle is bisected. Notice that the base of the isosceles triangle has length <math>a-(280-a)=2a-280</math>, so in the diagram above, <math>AE=280-a+\frac{1}{2}(2a-280)=280-a+a-140=140</math>.
 
Looking above at the right triangle containing <math>\theta</math>, we see that <math>\cos\theta=\frac{AE}{180}=\frac{140}{180}=\frac{7}{9}</math>. Hence <math>\left\lfloor 1000\cdot\frac{7}{9}\right\rfloor=\boxed{777}</math> is our answer.
 
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]


===Solution 3===
===Solution 3===

Latest revision as of 14:56, 25 October 2025

Problem

In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$. Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$)

Solution

Solution 1

[asy] real x = 1.60; /* arbitrary */ pointpen = black; pathpen = black+linewidth(0.7); size(180); real BD = x*x + 1.80*1.80 - 2 * 1.80 * x * 7 / 9; pair A=(0,0),B=(1.8,0),D=IP(CR(A,x),CR(B,BD)),C=OP(CR(D,1.8),CR(B,2.80 - x)); D(MP("A",A)--MP("B",B)--MP("C",C)--MP("D",D,N)--B--A--D); MP("180",(A+B)/2); MP("180",(C+D)/2,NE); D(anglemark(B,A,D)); D(anglemark(D,C,B)); [/asy]

By the Law of Cosines on $\triangle ABD$ at angle $A$ and on $\triangle BCD$ at angle $C$ (note $\angle C = \angle A$),

\[180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A\] \[(AD^2 - BC^2) = 360(AD - BC) \cos A\] \[(AD - BC)(AD + BC) = 360(AD - BC) \cos A\] \[(AD + BC) = 360 \cos A\] We know that $AD + BC = 640 - 360 = 280$. $\cos A = \dfrac{280}{360} = \dfrac{7}{9} = 0.777 \ldots$

$\lfloor 1000 \cos A \rfloor = \boxed{777}$.

Solution 2

Notice that $AB = CD$, and $BD = DB$, and $\angle{DAB} \cong \angle{BCD}$, so we have side-side-angle matching on triangles $ABD$ and $CDB$. Since the problem does not allow $\triangle{ABD} \cong \triangle{CDB}$, we know that $\angle{ADB}$ is not a right angle, and there is a unique other triangle with the matching side-side-angle.

[asy] pair A,B,C,D,F; A=(0,0);B=(5,4.04061);C=(6,0);D=(4,0);F=(5,0); draw(B--A--C--B--D); label("A",A,SW,red);label("E",F,S);label("B",B,NW,red);label("D",D,S,red);label("B",C,SE,blue);label("C",A,NW,blue);label("D",B,NE,blue);label("$\theta$",A,(4.5,1.5)); draw(anglemark(D,A,B,20));label("180",(A+B)/2,NW); draw(B--F,dashed);draw(rightanglemark(B,F,A));draw(D--B--C,dashed); add(pathticks(B--C,1,0.5,1,10));add(pathticks(B--D,1,0.5,1,10)); [/asy]

Overlay the triangles $\triangle BDA$ and $\triangle BDC$ on each other as in the diagram above (where the red labels correspond to $\triangle BDA$ and the blue labels correspond to $\triangle BDC$). Here we assume without loss of generality that $BC>AD$. Furthermore, let $\theta$ be the angle $A$ referenced in the problem; we need to find $\lfloor 1000\cos\theta\rfloor$.

Since the perimeter of $ABCD$ is $640$, we have $AD+BC=640-360=280$. Thus let $BC=a$ and $AD=280-a$ for some positive real number $a$. But the sides that correspond to $\overline{BD}$ above are congruent, so we can drop a perpendicular from the topmost point to a point $E$, where the base of the isosceles triangle is bisected. Notice that the base of the isosceles triangle has length $a-(280-a)=2a-280$, so in the diagram above, $AE=280-a+\frac{1}{2}(2a-280)=280-a+a-140=140$.

Looking above at the right triangle containing $\theta$, we see that $\cos\theta=\frac{AE}{180}=\frac{140}{180}=\frac{7}{9}$. Hence $\left\lfloor 1000\cdot\frac{7}{9}\right\rfloor=\boxed{777}$ is our answer.

~ eevee9406

Solution 3

Start the same as solution 1. We get \[180^2+a^2-360a \cos A = 180^2+b^2-360b \cos A \Rightarrow a^2-360a \cos A = b^2-360b \cos A,\]where $a$ is the length of $BC$ and $b$ is the length of $AD$. Let the common value of $a^2-360a \cos A$ and $b^2-360b \cos A$ be $c$. Then, the quadratic in $x$ \[x^2-360 \cos A \cdot x - c = 0\]has solutions $a$ and $b$. Therefore, by Vieta's, $a+b = 360 \cos A$. However, we know that the perimeter of $ABCD$ is $640$, so $a+b+180+180=640$, so $a+b=280$. Therefore, \[360 \cos A = 280 \Rightarrow \cos A = \frac{280}{360} \Rightarrow \cos A = \frac{7}{9} \Rightarrow \lfloor 1000 \cos A \rfloor = \boxed{777}.\]

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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