Difference between revisions of "1992 AIME Problems/Problem 9"
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Trapezoid <math>ABCD^{}_{}</math> has sides <math>AB=92^{}_{}</math>, <math>BC=50^{}_{}</math>, <math>CD=19^{}_{}</math>, and <math>AD=70^{}_{}</math>, with <math>AB^{}_{}</math> parallel to <math>CD^{}_{}</math>. A circle with center <math>P^{}_{}</math> on <math>AB^{}_{}</math> is drawn tangent to <math>BC^{}_{}</math> and <math>AD^{}_{}</math>. Given that <math>AP^{}_{}=\frac mn</math>, where <math>m^{}_{}</math> and <math>n^{}_{}</math> are relatively prime positive integers, find <math>m+n^{}_{}</math>. | Trapezoid <math>ABCD^{}_{}</math> has sides <math>AB=92^{}_{}</math>, <math>BC=50^{}_{}</math>, <math>CD=19^{}_{}</math>, and <math>AD=70^{}_{}</math>, with <math>AB^{}_{}</math> parallel to <math>CD^{}_{}</math>. A circle with center <math>P^{}_{}</math> on <math>AB^{}_{}</math> is drawn tangent to <math>BC^{}_{}</math> and <math>AD^{}_{}</math>. Given that <math>AP^{}_{}=\frac mn</math>, where <math>m^{}_{}</math> and <math>n^{}_{}</math> are relatively prime positive integers, find <math>m+n^{}_{}</math>. | ||
− | == Solution 1== | + | ==Solution 1== |
+ | Let <math>AP=x</math> so that <math>PB=92-x.</math> Extend <math>AD, BC</math> to meet at <math>X,</math> and note that <math>XP</math> bisects <math>\angle AXB;</math> let it meet <math>CD</math> at <math>E.</math> Using the angle bisector theorem, we let <math>XB=y(92-x), XA=xy</math> for some <math>y.</math> | ||
+ | |||
+ | Then <math>XD=xy-70, XC=y(92-x)-50,</math> thus <cmath>\frac{xy-70}{y(92-x)-50} = \frac{XD}{XC} = \frac{ED}{EC}=\frac{AP}{PB} = \frac{x}{92-x},</cmath> which we can rearrange, expand and cancel to get <math>120x=70\cdot 92,</math> hence <math>AP=x=\frac{161}{3}</math>. This gives us a final answer of <math>161+3=\boxed{164}</math> | ||
+ | |||
+ | == Solution 2== | ||
Let <math>AB</math> be the base of the trapezoid and consider angles <math>A</math> and <math>B</math>. Let <math>x=AP</math> and let <math>h</math> equal the height of the trapezoid. Let <math>r</math> equal the radius of the circle. | Let <math>AB</math> be the base of the trapezoid and consider angles <math>A</math> and <math>B</math>. Let <math>x=AP</math> and let <math>h</math> equal the height of the trapezoid. Let <math>r</math> equal the radius of the circle. | ||
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 and 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>. |
− | + | == Solution 3 == | |
− | |||
− | |||
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> | ||
− | + | == Solution 4 == | |
− | |||
− | |||
− | |||
− | |||
− | == Solution | ||
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{ | + | \frac{AD}{AP} &= \frac{BC}{BP}\\ |
&=\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>. | ||
+ | |||
+ | Note: this solution shows that the length of <math>CD</math> is irrelevant as long as there still exists a circle as described in the problem. | ||
+ | |||
+ | ==Solution 5== | ||
+ | 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 <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>. | ||
+ | |||
+ | <math>[BPC] = \frac{1}{2} \cdot 50 \cdot r</math> (where <math>r</math> is the radius of the tangent circle.) | ||
+ | |||
+ | <math>[CPD] = \frac{1}{2} \cdot 19 \cdot h</math> | ||
+ | |||
+ | <math>[PBA] = \frac{1}{2} \cdot 70 \cdot r</math> | ||
+ | |||
+ | <math>[BPC] + [CPD] + [PBA] = 60r + \frac{19h}{2} = [ABCD] = \frac{(19+92)h}{2}</math> | ||
+ | |||
+ | <math>60r = 46h</math> | ||
+ | |||
+ | <math>r = \frac{23h}{30}</math> | ||
+ | |||
+ | From Solution 1 above, <math>\frac{h}{70} = \frac{r}{x}</math> | ||
+ | |||
+ | Substituting <math>r = \frac{23h}{30}</math>, we find <math>x = \frac{161}{3}</math>, hence the answer is <math>\boxed{164}</math>. | ||
+ | ==Solution 6== | ||
+ | As the problem tells, the circle is tangent to both sides <math>AD,BC</math>, we can make it up to a triangle <math>QAB</math> and <math>P</math> must lie on its angular bisector. Then we know that <math>AP:BP=7:5</math>, which makes <math>AP = \frac{7}{12}(AB) = \frac{92\times 7}{12} = \frac{161}{3}</math>. Thus, <math>m + n = 164</math>. | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:27, 22 August 2021
Contents
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 so that
Extend
to meet at
and note that
bisects
let it meet
at
Using the angle bisector theorem, we let
for some
Then thus
which we can rearrange, expand and cancel to get
hence
. This gives us a final answer of
Solution 2
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 3
From above,
and
. Adding these equations yields
. Thus,
, and
.
We can use from Solution 1 to find that
and
.
This implies that so
Solution 4
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,
.
Note: this solution shows that the length of is irrelevant as long as there still exists a circle as described in the problem.
Solution 5
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
.
Solution 6
As the problem tells, the circle is tangent to both sides , we can make it up to a triangle
and
must lie on its angular bisector. Then we know that
, which makes
. Thus,
.
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 |
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