Difference between revisions of "2008 AMC 12B Problems/Problem 25"
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Multiplying this by 12, we find that the area of hexagon <math>ABQCDP</math> is <math>30\sqrt{3}</math>, which corresponds to answer choice <math>\boxed{B}</math>. | Multiplying this by 12, we find that the area of hexagon <math>ABQCDP</math> is <math>30\sqrt{3}</math>, which corresponds to answer choice <math>\boxed{B}</math>. | ||
+ | ==Solution 3== | ||
+ | <asy> | ||
+ | unitsize(0.6cm); | ||
+ | import olympiad; | ||
+ | pair A,B,C,D,P,Q,M,N,W,X,Y,Z; | ||
+ | A=(11/2,5sqrt(3)/2); | ||
+ | B=(33/2,5sqrt(3)/2); | ||
+ | C=(19,0); | ||
+ | D=(0,0); | ||
+ | P=incenter(A,D,(99999,5sqrt(3)/4)); | ||
+ | Q=incenter(B,C,(-99999,5sqrt(3)/4)); | ||
+ | W=P+(0,5sqrt(3)/4); | ||
+ | X=P-(0,5sqrt(3)/4); | ||
+ | Y=Q+(0,5sqrt(3)/4); | ||
+ | Z=Q-(0,5sqrt(3)/4); | ||
+ | M=reflect(A,P)*W; | ||
+ | N=reflect(B,Q)*Y; | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(A--P--D); | ||
+ | draw(B--Q--C); | ||
+ | label("$A$",A,dir(135)); | ||
+ | label("$B$",B,dir(45)); | ||
+ | label("$C$",C,dir(315)); | ||
+ | label("$D$",D,dir(225)); | ||
+ | dot("$P$",P,dir(0)); | ||
+ | dot("$Q$",Q,dir(180)); | ||
+ | draw(W--X); | ||
+ | draw(Y--Z); | ||
+ | draw(M--P); | ||
+ | draw(N--Q); | ||
+ | label("$11$",midpoint(A--B),dir(90)); | ||
+ | label("$5$",midpoint(B--C),dir(45)); | ||
+ | label("$19$",midpoint(C--D),dir(270)); | ||
+ | label("$7$",midpoint(D--A),dir(135)); | ||
+ | label("$x$",midpoint(P--W),dir(0)); | ||
+ | label("$x$",midpoint(P--X),dir(0)); | ||
+ | label("$x$",midpoint(P--M),dir(225)); | ||
+ | label("$x$",midpoint(Q--Y),dir(180)); | ||
+ | label("$x$",midpoint(Q--Z),dir(180)); | ||
+ | label("$x$",midpoint(Q--N),dir(315)); | ||
+ | draw(rightanglemark(P,W,B,12.5)); | ||
+ | draw(rightanglemark(P,X,C,12.5)); | ||
+ | draw(rightanglemark(P,M,D,12.5)); | ||
+ | draw(rightanglemark(Q,Y,A,12.5)); | ||
+ | draw(rightanglemark(Q,Z,D,12.5)); | ||
+ | draw(rightanglemark(Q,N,C,12.5)); | ||
+ | </asy> | ||
+ | |||
+ | Since point <math>P</math> is the intersection of the angle bisectors of <math>\angle A</math> and <math>\angle D</math>, <math>P</math> is equidistant from <math>\overline{AB}</math>, <math>\overline{AD}</math>, and <math>\overline{CD}</math>. Likewise, point <math>Q</math> is equidistant from <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CD}</math>. Because both points <math>P</math> and <math>Q</math> are equidistant from <math>\overline{AB}</math> and <math>\overline{CD}</math> and the distance between <math>\overline{AB}</math> and <math>\overline{CD}</math> is constant, the common distances from each of the points to the mentioned segments is equal for <math>P</math> and <math>Q</math>. Call this distance <math>x</math>. | ||
+ | |||
+ | The distance between a point and a line is the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from <math>P</math> to <math>\overline{AD}</math> is <math>x</math>, so the area of <math>\triangle ADP</math> is equal to <math>\frac12\cdot AD\cdot x=\frac72x</math>. Similarly, the area of <math>\triangle BCQ</math> is <math>\frac12\cdot BC\cdot x=\frac52x</math>. The altitude of the trapezoid is <math>2x</math>, because it is the sum of the distances from either <math>P</math> or <math>Q</math> to <math>\overline{AB}</math> and <math>\overline{CD}</math>. This means the area of trapezoid <math>ABCD</math> is <math>\frac12(AB+CD)\cdot2x=\frac12(11+19)\cdot2x=30x</math>. Now, the area of hexagon <math>ABQCDP</math> is the area of trapezoid <math>ABCD</math>, minus the areas of triangles <math>ADP</math> and <math>BCQ</math>. This is <math>30x-\frac72x-\frac52x=24x</math>. Now it remains to find <math>x</math>. | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.6cm); | ||
+ | import olympiad; | ||
+ | pair A,B,C,D,R,S; | ||
+ | A=(11/2,5sqrt(3)/2); | ||
+ | B=(33/2,5sqrt(3)/2); | ||
+ | C=(19,0); | ||
+ | D=(0,0); | ||
+ | R=(11/2,0); | ||
+ | S=(33/2,0); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(A--R); | ||
+ | draw(B--S); | ||
+ | label("$A$",A,dir(135)); | ||
+ | label("$B$",B,dir(45)); | ||
+ | label("$C$",C,dir(315)); | ||
+ | label("$D$",D,dir(225)); | ||
+ | label("$R$",R,dir(270)); | ||
+ | label("$S$",S,dir(270)); | ||
+ | label("$11$",midpoint(A--B),dir(90)); | ||
+ | label("$5$",midpoint(B--C),dir(45)); | ||
+ | label("$11$",midpoint(R--S),dir(270)); | ||
+ | label("$7$",midpoint(D--A),dir(135)); | ||
+ | label("$r$",midpoint(R--D),dir(270)); | ||
+ | label("$s$",midpoint(C--S),dir(270)); | ||
+ | label("$19$",midpoint(C--D),5*dir(270)); | ||
+ | label("$2x$",midpoint(A--R),dir(0)); | ||
+ | label("$2x$",midpoint(B--S),dir(180)); | ||
+ | draw(rightanglemark(A,R,D,15)); | ||
+ | draw(rightanglemark(B,S,C,15)); | ||
+ | </asy> | ||
+ | |||
+ | We let <math>R</math> and <math>S</math> be the feet of the altitudes of <math>A</math> and <math>B</math>, respectively, to <math>\overline{CD}</math>. We define <math>r=RD</math> and <math>s=SC</math>. We know that <math>AB=RS</math>, so <math>RS=11</math> and <math>r+s=19-11=8</math>. By the Pythagorean Theorem on <math>\triangle ADR</math> and <math>\triangle BCS</math>, we get <math>r^2+(2x)^2=7^2</math> and <math>s^2+(2x)^2=5^2</math>, respectively. Subtracting the second equation from the first gives us <math>r^2-s^2=49-25=24</math>. The left hand side of this equation is a difference of squares and factors to <math>(r+s)(r-s)</math>. We know that <math>r+s=8</math>, so <math>r-s=\frac{24}8=3</math>. Now we can solve for <math>r</math> by adding the two equations we just got to see that <math>2r=11</math>, or <math>r=\frac{11}2</math>. | ||
+ | |||
+ | We now solve for <math>x</math>. We know that <math>r^2+(2x)^2=49</math>, so <math>(2x)^2=49-\left(\frac{11}2\right)^2=\frac{75}4</math> and <math>2x=\frac{5\sqrt3}2</math>. We multiply both sides of this equation by <math>12</math> to get <math>24x=30\sqrt3</math>. However, the area of hexagon <math>ABQCDP</math> is <math>24x</math>, so the answer is <math>30\sqrt 3</math>, or answer choice <math>\boxed{B}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=24|after=Last Question}} | {{AMC12 box|year=2008|ab=B|num-b=24|after=Last Question}} |
Revision as of 16:16, 6 April 2018
Problem 25
Let be a trapezoid with and . Bisectors of and meet at , and bisectors of and meet at . What is the area of hexagon ?
Solution
Note: In the image AB and CD have been swapped from their given lengths in the problem. However, this doesn't affect any of the solving.
Drop perpendiculars to from and , and call the intersections respectively. Now, and . Thus, . We conclude and . To simplify things even more, notice that , so .
Also, So the area of is:
Over to the other side: is , and is therefore congruent to . So .
The area of the hexagon is clearly
Alternate Solution
Let and meet at and , respectively.
Since , , and they share , triangles and are congruent.
By the same reasoning, we also have that triangles and are congruent.
Hence, we have .
If we let the height of the trapezoid be , we have .
Thusly, if we find the height of the trapezoid and multiply it by 12, we will be done.
Let the projections of and to be and , respectively.
We have , , and .
Therefore, . Solving this, we easily get that .
Multiplying this by 12, we find that the area of hexagon is , which corresponds to answer choice .
Solution 3
Since point is the intersection of the angle bisectors of and , is equidistant from , , and . Likewise, point is equidistant from , , and . Because both points and are equidistant from and and the distance between and is constant, the common distances from each of the points to the mentioned segments is equal for and . Call this distance .
The distance between a point and a line is the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from to is , so the area of is equal to . Similarly, the area of is . The altitude of the trapezoid is , because it is the sum of the distances from either or to and . This means the area of trapezoid is . Now, the area of hexagon is the area of trapezoid , minus the areas of triangles and . This is . Now it remains to find .
We let and be the feet of the altitudes of and , respectively, to . We define and . We know that , so and . By the Pythagorean Theorem on and , we get and , respectively. Subtracting the second equation from the first gives us . The left hand side of this equation is a difference of squares and factors to . We know that , so . Now we can solve for by adding the two equations we just got to see that , or .
We now solve for . We know that , so and . We multiply both sides of this equation by to get . However, the area of hexagon is , so the answer is , or answer choice .
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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