Difference between revisions of "1992 AIME Problems/Problem 9"
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Then | Then | ||
− | < | + | <cmath>\sin{A}= \frac{r}{x} = \frac{h}{70}\qquad\text{ and }\qquad\sin{B}= \frac{r}{92-x} = \frac{h}{50}.\tag{1}</cmath> |
Let <math>z</math> be the distance along <math>AB</math> from <math>A</math> to where the perp from <math>D</math> meets <math>AB</math>. | Let <math>z</math> be the distance along <math>AB</math> from <math>A</math> to where the perp from <math>D</math> meets <math>AB</math>. | ||
− | Then <math>h^2 +z^2 =70^2</math> and <math>(73-z)^2 + h^2 =50^2</math> so <math>h =\frac{\sqrt{44710959}}{146}</math> | + | Then <math>h^2 +z^2 =70^2</math> and <math>(73-z)^2 + h^2 =50^2</math> so <math>h =\frac{\sqrt{44710959}}{146}</math>. |
− | + | We can substitute this into <math>(1)</math> to find that <math>x= \frac{11753}{219} = \frac{161}{3}</math> and <math>m+n = 164</math>. | |
− | + | <b>Remark:</b> One can come up with the equations in <math>(1)</math> without directly resorting to trig. From similar triangles, | |
− | <math>h/r = 70/x</math> and <math>h/r = 50/ (92-x)</math> | + | <math>h/r = 70/x</math> and <math>h/r = 50/ (92-x)</math>. This implies that <math>70/x =50/(92-x)</math>, so <math>x = 161/3</math>. |
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== Solution 2 == | == Solution 2 == | ||
From <math>(1)</math> above, <math>x = \frac{70r}{h}</math> and <math>92-x = \frac{50r}{h}</math>. Adding these equations yields <math>92 = \frac{120r}{h}</math>. Thus, <math>x = \frac{70r}{h} = \frac{7}{12}\cdot\frac{120r}{h} = \frac{7}{12}\cdot92 = \frac{161}{3}</math>, and <math>m+n = \boxed{164}</math>. | From <math>(1)</math> above, <math>x = \frac{70r}{h}</math> and <math>92-x = \frac{50r}{h}</math>. Adding these equations yields <math>92 = \frac{120r}{h}</math>. Thus, <math>x = \frac{70r}{h} = \frac{7}{12}\cdot\frac{120r}{h} = \frac{7}{12}\cdot92 = \frac{161}{3}</math>, and <math>m+n = \boxed{164}</math>. | ||
+ | We can use <math>(1)</math> from Solution 1 to find that <math>h/r = 70/x</math> and <math>h/r = 50/ (92-x)</math>. | ||
− | + | This implies that <math>70/x =50/(92-x)</math> so <math>x = 161/3</math> | |
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== Solution 3 == | == Solution 3 == | ||
Extend <math>AD</math> and <math>BC</math> to meet at a point <math>X</math>. Since <math>AB</math> and <math>CD</math> are parallel, <math>\triangle XCD ~ \triangle XAB</math>. If <math>AX</math> is further extended to a point <math>A'</math> and <math>XB</math> is extended to a point <math>B'</math> such that <math>A'B'</math> is tangent to circle <math>P</math>, we discover that circle <math>P</math> is the incircle of triangle <math>XA'B'</math>. Then line <math>XP</math> is the angle bisector of <math>\angle AXB</math>. By homothety, <math>P</math> is the intersection of the angle bisector of <math>\triangle XAB</math> with <math>AB</math>. By the angle bisector theorem, | Extend <math>AD</math> and <math>BC</math> to meet at a point <math>X</math>. Since <math>AB</math> and <math>CD</math> are parallel, <math>\triangle XCD ~ \triangle XAB</math>. If <math>AX</math> is further extended to a point <math>A'</math> and <math>XB</math> is extended to a point <math>B'</math> such that <math>A'B'</math> is tangent to circle <math>P</math>, we discover that circle <math>P</math> is the incircle of triangle <math>XA'B'</math>. Then line <math>XP</math> is the angle bisector of <math>\angle AXB</math>. By homothety, <math>P</math> is the intersection of the angle bisector of <math>\triangle XAB</math> with <math>AB</math>. By the angle bisector theorem, | ||
− | < | + | <cmath>\begin{align*} |
\frac{AX}{AP} &= \frac{XB}{BP}\\ | \frac{AX}{AP} &= \frac{XB}{BP}\\ | ||
\frac{AX}{AP} - \frac{XD}{AP} &= \frac{XB}{BP} - \frac{XC}{BP}\\ | \frac{AX}{AP} - \frac{XD}{AP} &= \frac{XB}{BP} - \frac{XC}{BP}\\ | ||
\frac{AD}{AP} &= \frac{BD}{PB}\\ | \frac{AD}{AP} &= \frac{BD}{PB}\\ | ||
&=\frac{7}{5} | &=\frac{7}{5} | ||
− | \end{align*}</ | + | \end{align*}</cmath> |
Let <math>7a = AP</math>, then <math>AB = 7a + 5a = 12a</math>. <math>AP = \frac{7}{12}(AB) = \frac{92\times 7}{12} = \frac{161}{3}</math>. Thus, <math>m + n = 164</math>. | Let <math>7a = AP</math>, then <math>AB = 7a + 5a = 12a</math>. <math>AP = \frac{7}{12}(AB) = \frac{92\times 7}{12} = \frac{161}{3}</math>. Thus, <math>m + n = 164</math>. | ||
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The area of the trapezoid is <math>\frac{(19+92)h}{2}</math>, where <math>h</math> is the height of the trapezoid. | The area of the trapezoid is <math>\frac{(19+92)h}{2}</math>, where <math>h</math> is the height of the trapezoid. | ||
− | Draw lines CP and BP. We can now find the area of the trapezoid as the sum of the three triangles BPC, CPD, and PBA. | + | Draw lines <math>CP</math> and <math>BP</math>. We can now find the area of the trapezoid as the sum of the areas of the three triangles <math>BPC</math>, <math>CPD</math>, and <math>PBA</math>. |
− | [BPC] = | + | <math>[BPC] = \frac{1}{2} \cdot 50 \cdot r</math> (where <math>r</math> is the radius of the tangent circle.) |
− | [CPD] = | + | <math>[CPD] = \frac{1}{2} \cdot 19 \cdot h</math> |
− | [PBA] = | + | <math>[PBA] = \frac{1}{2} \cdot 70 \cdot r</math> |
− | [BPC] + [CPD] + [PBA] = | + | <math>[BPC] + [CPD] + [PBA] = 60r + \frac{19h}{2} = [ABCD] = \frac{(19+92)h}{2}</math> |
<math>60r = 46h</math> | <math>60r = 46h</math> | ||
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From Solution 1 above, <math>\frac{h}{70} = \frac{r}{x}</math> | From Solution 1 above, <math>\frac{h}{70} = \frac{r}{x}</math> | ||
− | Substituting <math>r = \frac{23h}{30}</math>, we | + | Substituting <math>r = \frac{23h}{30}</math>, we find <math>x = \frac{161}{3}</math>, hence the answer is <math>\boxed{164}</math>. |
== See also == | == See also == |
Revision as of 16:25, 13 March 2015
Problem
Trapezoid has sides
,
,
, and
, with
parallel to
. A circle with center
on
is drawn tangent to
and
. Given that
, where
and
are relatively prime positive integers, find
.
Solution 1
Let be the base of the trapezoid and consider angles
and
. Let
and let
equal the height of the trapezoid. Let
equal the radius of the circle.
Then
Let be the distance along
from
to where the perp from
meets
.
Then and
so
.
We can substitute this into
to find that
and
.
Remark: One can come up with the equations in without directly resorting to trig. From similar triangles,
and
. This implies that
, so
.
Solution 2
From above,
and
. Adding these equations yields
. Thus,
, and
.
We can use from Solution 1 to find that
and
.
This implies that so
Solution 3
Extend and
to meet at a point
. Since
and
are parallel,
. If
is further extended to a point
and
is extended to a point
such that
is tangent to circle
, we discover that circle
is the incircle of triangle
. Then line
is the angle bisector of
. By homothety,
is the intersection of the angle bisector of
with
. By the angle bisector theorem,
Let , then
.
. Thus,
.
Solution 4
The area of the trapezoid is , where
is the height of the trapezoid.
Draw lines and
. We can now find the area of the trapezoid as the sum of the areas of the three triangles
,
, and
.
(where
is the radius of the tangent circle.)
From Solution 1 above,
Substituting , we find
, hence the answer is
.
See also
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.