Difference between revisions of "2008 AIME II Problems/Problem 5"
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− | == Problem | + | __TOC__ |
+ | |||
+ | == Problem== | ||
In [[trapezoid]] <math>ABCD</math> with <math>\overline{BC}\parallel\overline{AD}</math>, let <math>BC = 1000</math> and <math>AD = 2008</math>. Let <math>\angle A = 37^\circ</math>, <math>\angle D = 53^\circ</math>, and <math>M</math> and <math>N</math> be the [[midpoint]]s of <math>\overline{BC}</math> and <math>\overline{AD}</math>, respectively. Find the length <math>MN</math>. | In [[trapezoid]] <math>ABCD</math> with <math>\overline{BC}\parallel\overline{AD}</math>, let <math>BC = 1000</math> and <math>AD = 2008</math>. Let <math>\angle A = 37^\circ</math>, <math>\angle D = 53^\circ</math>, and <math>M</math> and <math>N</math> be the [[midpoint]]s of <math>\overline{BC}</math> and <math>\overline{AD}</math>, respectively. Find the length <math>MN</math>. | ||
− | |||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | Extend <math>\overline{ | + | Extend <math>\overline{AB}</math> and <math>\overline{CD}</math> to meet at a point <math>E</math>. Then <math>\angle AED = 180 - 53 - 37 = 90^{\circ}</math>. |
<center><asy> | <center><asy> | ||
size(220); | size(220); | ||
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label("\(500\)",(M[0]+C)/2,S); | label("\(500\)",(M[0]+C)/2,S); | ||
</asy></center> | </asy></center> | ||
− | + | As <math>\angle AED = 90^{\circ}</math>, note that the midpoint of <math>\overline{AD}</math>, <math>N</math>, is the center of the [[circumcircle]] of <math>\triangle AED</math>. We can do the same with the circumcircle about <math>\triangle BEC</math> and <math>M</math> (or we could apply the homothety to find <math>ME</math> in terms of <math>NE</math>). It follows that | |
+ | |||
+ | <cmath>NE = ND = \frac {AD}{2} = 1004, \quad ME = MC = \frac {BC}{2} = 500.</cmath> | ||
− | |||
− | |||
− | |||
− | |||
Thus <math>MN = NE - ME = \boxed{504}</math>. | Thus <math>MN = NE - ME = \boxed{504}</math>. | ||
+ | |||
+ | |||
+ | For purposes of rigor we will show that <math>E,M,N</math> are collinear. Since <math>\overline{BC} \parallel \overline{AD}</math>, then <math>BC</math> and <math>AD</math> are [[homothecy|homothetic]] with respect to point <math>E</math> by a ratio of <math>\frac{BC}{AD} = \frac{125}{251}</math>. Since the homothety carries the midpoint of <math>\overline{BC}</math>, <math>M</math>, to the midpoint of <math>\overline{AD}</math>, which is <math>N</math>, then <math>E,M,N</math> are [[collinear]]. | ||
=== Solution 2 === | === Solution 2 === | ||
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label("\(504-x\)",(G+D)/2,S); | label("\(504-x\)",(G+D)/2,S); | ||
label("\(504+x\)",(A+F)/2,S); | label("\(504+x\)",(A+F)/2,S); | ||
− | label("\(h\)",(M+H)/2, | + | label("\(h\)",(M[0]+H)/2,(1,0)); |
</asy></center> | </asy></center> | ||
Let <math>F,G,H</math> be the feet of the [[perpendicular]]s from <math>B,C,M</math> onto <math>\overline{AD}</math>, respectively. Let <math>x = NH</math>, so <math>DG = 1004 - 500 - x = 504 - x</math> and <math>AF = 1004 - (500 - x) = 504 + x</math>. Also, let <math>h = BF = CG = HM</math>. | Let <math>F,G,H</math> be the feet of the [[perpendicular]]s from <math>B,C,M</math> onto <math>\overline{AD}</math>, respectively. Let <math>x = NH</math>, so <math>DG = 1004 - 500 - x = 504 - x</math> and <math>AF = 1004 - (500 - x) = 504 + x</math>. Also, let <math>h = BF = CG = HM</math>. | ||
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=== Solution 3 === | === Solution 3 === | ||
− | If you drop perpendiculars from B and C to AD, | + | If you drop perpendiculars from <math>B</math> and <math>C</math> to <math>AD</math>, and call the points where they meet <math>\overline{AD}</math>, <math>E</math> and <math>F</math> respectively, then <math>FD = x</math> and <math>EA = 1008-x</math> , and so you can solve an equation in tangents. Since <math>\angle{A} = 37</math> and <math>\angle{D} = 53</math>, you can solve the equation [by cross-multiplication]: |
− | < | + | <cmath>\begin{align*}\tan{37}\times (1008-x) &= \tan{53} \times x\\ |
+ | \frac{(1008-x)}{x} &= \frac{\tan{53}}{\tan{37}} = \frac{\sin{53}}{\cos{53}} \times\frac{\sin{37}}{\cos{37}}\end{align*}</cmath> | ||
− | + | However, we know that <math>\cos{90-x} = \sin{x}</math> and <math>\sin{90-x} = \cos{x}</math> are co-functions. Applying this, | |
− | < | + | <cmath>\begin{align*}\frac{(1008-x)}{x} &= \frac{\sin^2{53}}{\cos^2{53}} \\ |
+ | x\sin^2{53} &= 1008\cos^2{53} - x\cos^2{53}\\ | ||
+ | x(\sin^2{53} + \cos^2{53}) &= 1008\cos^2{53}\\ | ||
+ | x = 1008\cos^2{53} &\Longrightarrow 1008-x = 1008\sin^2{53} | ||
+ | \end{align*}</cmath> | ||
+ | Now, if we can find <math>1004 - (EA + 500)</math>, and the height of the trapezoid, we can create a right triangle and use the [[Pythagorean Theorem]] to find <math>MN</math>. | ||
− | + | The leg of the right triangle along the horizontal is: | |
− | < | + | <cmath>1004 - 1008\sin^2{53} - 500 = 504 - 1008\sin^2{53}.</cmath> |
− | + | Now to find the other leg of the right triangle (also the height of the trapezoid), we can simplify the following expression: | |
− | < | + | <cmath>\begin{align*}\tan{37} \times 1008 \sin^2{53} |
− | + | = \tan{37} \times 1008 \cos^2{37} | |
− | + | = 1008\cos{37}\sin{37} | |
− | + | = 504\sin74\end{align*}</cmath> | |
− | Now | + | Now we used Pythagorean Theorem and get that <math>MN</math> is equal to: |
− | + | <cmath> | |
+ | \begin{align*}&\sqrt{(1008\sin^2{53} + 500 -1004)^2 + (504\sin{74})^2} = 504\sqrt{1-2\sin^2{53} + \sin^2{74}} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
− | <math> | + | However, <math>1-2\sin^2{53} = \cos^2{106}</math> and <math>\sin^2{74} = \sin^2{106}</math> so now we end up with: |
− | + | <cmath>504\sqrt{\cos^2{106} + \sin^2{106}} =\fbox{504}.</cmath> | |
− | + | === Solution 4 === | |
− | = | ||
− | = | ||
− | = | ||
− | + | Plot the trapezoid such that <math>B=\left(1000\cos 37^\circ, 0\right)</math>, <math>C=\left(0, 1000\sin 37^\circ\right)</math>, <math>A=\left(2008\cos 37^\circ, 0\right)</math>, and <math>D=\left(0, 2008\sin 37^\circ\right)</math>. | |
− | <math>\ | + | The midpoints of the requested sides are <math>\left(500\cos 37^\circ, 500\sin 37^\circ\right)</math> and <math>\left(1004\cos 37^\circ, 1004\sin 37^\circ\right)</math>. |
− | + | To find the distance from <math>M</math> to <math>N</math>, we simply apply the distance formula and the Pythagorean identity <math>\sin^2 x + \cos^2 x = 1</math> to get <math>MN=\boxed{504}</math>. | |
− | |||
− | |||
− | + | === Solution 5 === | |
− | + | Similar to solution 1; Notice that it forms a right triangle. Remembering that the median to the hypotenuse is simply half the length of the hypotenuse, we quickly see that the length is <math>\frac{2008}{2}-\frac{1000}{2}=504</math>. | |
− | + | === Solution 6 === | |
− | = | + | Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially <math>\boxed{504}</math>. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.) |
− | |||
+ | == Solution 7 == | ||
+ | Let the height be h. Note that if <math>\overline{NH} = x</math> then if we draw perpendiculars like in solution 1, <math>\overline{FN} = 500 - x, \overline{AF} = 504 + x, \overline{HG} = 500, \overline{GD} = 504 - x.</math> Note that we wish to find <math>\overline{MN} = \sqrt{x^2 + h^2}.</math> Let's find <math>\tan(53)</math> in two ways. Finding <math>\tan(53)</math> from <math>\triangle BAF</math> yields <math>\tan(53) = \frac{504+x}{h}.</math> Finding it from <math>\triangle CDG</math> yields <math>\frac{h}{504-x}.</math> Setting these equal yields | ||
+ | <cmath>\frac{504+x}{h}=\frac{h}{504-x} \rightarrow h^2 = 504^2-x^2 \rightarrow \sqrt{x^2+h^2} = \sqrt{504^2} = \boxed{504}</cmath> | ||
+ | == Solution 8 == | ||
+ | Rotate trapezoid <math>MNCD</math> 180 degrees around point <math>N</math> so that <math>AN</math> coincides with <math>ND</math>. Let the image of trapezoid <math>MNCD</math> be <math>ANM'C'</math>. Since angles are preserved during rotations, <math>\angle BAC' = 37^{\circ} + 53 ^{\circ} =90 ^{\circ}</math>. Since <math>BM=CM=C'M'</math> and <math>BM || C'M'</math>, <math>BMM'C'</math> is a parallelogram. Thus, <math>MM'=BC'</math>. | ||
+ | Let the point where <math>BC'</math> intersects <math>AD</math> be <math>E</math>. Since <math>BMNE</math> is a parallelogram, <math>AE=AN-BM=1004-500-504</math>. Since <math>BE=EC</math> and <math>\angle BAC= 90^{\circ}</math>, <math>AE</math> is a median to the hypotenuse of <math>BAC'</math>. Therefore, <math>BC'=2 AE= 1008</math>, and <math>BE=MN=\boxed{504}.</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2008|n=II|num-b=4|num-a=6}} | {{AIME box|year=2008|n=II|num-b=4|num-a=6}} | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:15, 28 November 2023
Contents
Problem
In trapezoid with
, let
and
. Let
,
, and
and
be the midpoints of
and
, respectively. Find the length
.
Solution
Solution 1
Extend and
to meet at a point
. Then
.
![[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); draw(A--B--C--D--cycle); draw(B--E--C,dashed); draw(M[0]--N); draw(N--E,dashed); draw(rightanglemark(D,E,A,2)); picture p = new picture; draw(p,Circle(N,r),dashed+linewidth(0.5)); clip(p,A--D--D+(0,20)--A+(0,20)--cycle); add(p); label("\(A\)",A,SW); label("\(B\)",B,NW); label("\(C\)",C,NE); label("\(D\)",D,SE); label("\(E\)",E,NE); label("\(M\)",M[0],SW); label("\(N\)",N,S); label("\(1004\)",(N+D)/2,S); label("\(500\)",(M[0]+C)/2,S); [/asy]](http://latex.artofproblemsolving.com/a/d/e/ade36a6d8b6d97f88722facc76700d34501ddae8.png)
As , note that the midpoint of
,
, is the center of the circumcircle of
. We can do the same with the circumcircle about
and
(or we could apply the homothety to find
in terms of
). It follows that
Thus .
For purposes of rigor we will show that are collinear. Since
, then
and
are homothetic with respect to point
by a ratio of
. Since the homothety carries the midpoint of
,
, to the midpoint of
, which is
, then
are collinear.
Solution 2
![[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label("\(A\)",A,SW); label("\(B\)",B,NW); label("\(C\)",C,NE); label("\(D\)",D,NE); label("\(F\)",F,S); label("\(G\)",G,SW); label("\(M\)",M[0],SW); label("\(N\)",N,S); label("\(H\)",H,S); label("\(x\)",(N+H)/2+(0,1),S); label("\(h\)",(B+F)/2,W); label("\(h\)",(C+G)/2,W); label("\(1000\)",(B+C)/2,NE); label("\(504-x\)",(G+D)/2,S); label("\(504+x\)",(A+F)/2,S); label("\(h\)",(M[0]+H)/2,(1,0)); [/asy]](http://latex.artofproblemsolving.com/e/d/2/ed2323da960926cb22458648f561a6953f91de51.png)
Let be the feet of the perpendiculars from
onto
, respectively. Let
, so
and
. Also, let
.
By AA~, we have that , and so
By the Pythagorean Theorem on ,
so
.
Solution 3
If you drop perpendiculars from and
to
, and call the points where they meet
,
and
respectively, then
and
, and so you can solve an equation in tangents. Since
and
, you can solve the equation [by cross-multiplication]:
However, we know that and
are co-functions. Applying this,
Now, if we can find
, and the height of the trapezoid, we can create a right triangle and use the Pythagorean Theorem to find
.
The leg of the right triangle along the horizontal is:
Now to find the other leg of the right triangle (also the height of the trapezoid), we can simplify the following expression:
Now we used Pythagorean Theorem and get that is equal to:
However, and
so now we end up with:
Solution 4
Plot the trapezoid such that ,
,
, and
.
The midpoints of the requested sides are and
.
To find the distance from to
, we simply apply the distance formula and the Pythagorean identity
to get
.
Solution 5
Similar to solution 1; Notice that it forms a right triangle. Remembering that the median to the hypotenuse is simply half the length of the hypotenuse, we quickly see that the length is .
Solution 6
Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially . (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)
Solution 7
Let the height be h. Note that if then if we draw perpendiculars like in solution 1,
Note that we wish to find
Let's find
in two ways. Finding
from
yields
Finding it from
yields
Setting these equal yields
Solution 8
Rotate trapezoid 180 degrees around point
so that
coincides with
. Let the image of trapezoid
be
. Since angles are preserved during rotations,
. Since
and
,
is a parallelogram. Thus,
.
Let the point where
intersects
be
. Since
is a parallelogram,
. Since
and
,
is a median to the hypotenuse of
. Therefore,
, and
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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