Difference between revisions of "2011 AIME I Problems/Problem 2"
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== Solution 1 == | == Solution 1 == | ||
| + | <!-- | ||
| + | <asy> | ||
| + | unitsize(0.5cm); | ||
| + | defaultpen(0.8); | ||
| + | pair A=(0,0), B=(12,0), C=(12,10), D=(0,10); | ||
| + | draw(A--B--C--D--cycle); | ||
| + | label("$A$",A,SW); | ||
| + | label("$B$",B,SE); | ||
| + | label("$C$",C,NE); | ||
| + | label("$D$",D,NW); | ||
| + | |||
| + | pair E=(3,5), F=(9,5); | ||
| + | dot("$E$",E,N); | ||
| + | dot("$F$",F,N); | ||
| + | pair G = extension(A, D, E, F); | ||
| + | pair H = extension(B, C, E, F); | ||
| + | draw(G--H); | ||
| + | dot("$G$",G,W); | ||
| + | dot("$H$", H,E); | ||
| + | </asy> | ||
| + | --> | ||
Let us call the point where <math>\overline{EF}</math> intersects <math>\overline{AD}</math> point <math>G</math>, and the point where <math>\overline{EF}</math> intersects <math>\overline{BC}</math> point <math>H</math>. Since angles <math>FHB</math> and <math>EGA</math> are both right angles, and angles <math>BEF</math> and <math>DFE</math> are congruent due to parallelism, right triangles <math>BHE</math> and <math>DGF</math> are similar. This implies that <math>\frac{BH}{GD} = \frac{9}{8}</math>. Since <math>BC=10</math>, <math>BH+GD=BH+HC=BC=10</math>. (<math>HC</math> is the same as <math>GD</math> because they are opposite sides of a rectangle.) Now, we have a system: | Let us call the point where <math>\overline{EF}</math> intersects <math>\overline{AD}</math> point <math>G</math>, and the point where <math>\overline{EF}</math> intersects <math>\overline{BC}</math> point <math>H</math>. Since angles <math>FHB</math> and <math>EGA</math> are both right angles, and angles <math>BEF</math> and <math>DFE</math> are congruent due to parallelism, right triangles <math>BHE</math> and <math>DGF</math> are similar. This implies that <math>\frac{BH}{GD} = \frac{9}{8}</math>. Since <math>BC=10</math>, <math>BH+GD=BH+HC=BC=10</math>. (<math>HC</math> is the same as <math>GD</math> because they are opposite sides of a rectangle.) Now, we have a system: | ||
Revision as of 08:07, 14 May 2011
Contents
Problem
In rectangle 
, 
and 
. Points 
and 
lie inside rectangle so that 
,
,
,
, and line 
intersects segment 
. The length 
can be expressed in the form 
, where 
,
, and 
are positive integers and is not divisible by the square of any prime. Find 
.
Solution 1
Let us call the point where 
intersects point 
, and the point where intersects 
point 
. Since angles 
and 
are both right angles, and angles 
and 
are congruent due to parallelism, right triangles 
and 
are similar. This implies that 
. Since , 
. (
is the same as 
because they are opposite sides of a rectangle.) Now, we have a system:
Solving this system (easiest by substitution), we get that:
Using the Pythagorean Theorem, we can solve for the remaining sides of the two right triangles:
and 
Notice that adding these two sides would give us twelve plus the overlap . This means that:
Since 
isn't divisible by any perfect square, our answer is:
Solution 2
Extend lines and 
to meet at point .
Draw the altitude 
from point to line 
extended.
In right 
, 
, , thus by Pythagoras Theorem we have:
Thus our answer is:
See also
| 2011 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
