Difference between revisions of "2008 AIME II Problems/Problem 5"
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=== Solution 3 === | === Solution 3 === | ||
− | If you drop perpendiculars from B and C to AD, and call the points if you drop perpendiculars from B and C to AD and call the points where they meet AD E and F respectively and call FD = x and | + | If you drop perpendiculars from <math>B</math> and <math>C</math> to <math>AD</math>, and call the points if you drop perpendiculars from <math>B</math> and <math>C</math> to <math>\overline{AD}</math> and call the points where they meet <math>\overline{AD}</math>, <math>E</math> and <math>F</math> respectively and call <math>FD = x</math> and <math>EA = 1008-x</math> , then 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>. | |
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− | Now, if we can find <math>1004 - (EA + 500)</math>, and the height of the trapezoid, we can create a right triangle and | ||
The leg of the right triangle along the horizontal is: | 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: | 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> |
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− | + | Now we used Pythagorean Theorem and get that <math>MN</math> is equal to: | |
− | <math> | ||
− | + | <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> | ||
− | so now we end up with: | + | 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> |
− | |||
− | = | ||
== See also == | == See also == |
Revision as of 12:06, 8 June 2008
Problem 5
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)
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.
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
.
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+H)/2,W); [/asy]](http://latex.artofproblemsolving.com/c/0/b/c0b6074f2ecbecac49db1848604ea72ef51447fd.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 if you drop perpendiculars from
and
to
and call the points where they meet
,
and
respectively and call
and
, then 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:
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 |