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. | ||
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<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>. | <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 | + | == 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>. | ||
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This implies that <math>70/x =50/(92-x)</math> so <math>x = 161/3</math> | This implies that <math>70/x =50/(92-x)</math> so <math>x = 161/3</math> | ||
− | == Solution | + | == Solution 4 == |
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, | ||
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\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*}</cmath> | \end{align*}</cmath> | ||
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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>. | ||
− | ==Solution | + | 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. | The area of the trapezoid is <math>\frac{(19+92)h}{2}</math>, where <math>h</math> is the height of the trapezoid. | ||
Latest revision as of 00:14, 23 February 2020
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 .
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.