Difference between revisions of "2008 AMC 12B Problems/Problem 25"

Revision as of 19:54, 9 October 2022

Problem

Let $ABCD$ be a trapezoid with $AB||CD, AB=11, BC=5, CD=19,$ and $DA=7$ . Bisectors of $\angle A$ and $\angle D$ meet at $P$ , and bisectors of $\angle B$ and $\angle C$ meet at $Q$ . What is the area of hexagon $ABQCDP$ ?

$\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textbf{(D)}\ 35\sqrt{3}\qquad \textbf{(E)}\ 36\sqrt{3}$

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 $CD$ from $A$ and $B$ , and call the intersections $X,Y$ respectively. Now, $DA^2-BC^2=(7-5)(7+5)=DX^2-CY^2$ and $DX+CY=19-11=8$ . Thus, $DX-CY=3$ . We conclude $DX=\frac{11}{2}$ and $CY=\frac{5}{2}$ . To simplify things even more, notice that $90^{\circ}=\frac{\angle D+\angle A}{2}=180^{\circ}-\angle APD$ , so $\angle P=\angle Q=90^{\circ}$ .

Also, $\sin(\angle PAD)=\sin(\frac12\angle XDA)=\sqrt{\frac{1-\cos(\angle XDA)}{2}}=\sqrt{\frac{3}{28}}$ So the area of $\triangle APD$ is: $R\cdot c\sin a\sin b =\frac{7\cdot7}{2}\sqrt{\frac{3}{28}}\sqrt{1-\frac{3}{28}}=\frac{35}{8}\sqrt{3}$

Over to the other side: $\triangle BCY$ is $30-60-90$ , and is therefore congruent to $\triangle BCQ$ . So $[BCQ]=\frac{5\cdot5\sqrt{3}}{8}$ .

The area of the hexagon is clearly $\begin{align*} [ABCD]-([BCQ]+[APD]) &=\frac{15\cdot5\sqrt{3}}{2}-\frac{60\sqrt{3}}{8}\\ &=30\sqrt{3}\implies\boxed{\mathrm{B}} \end{align*}$

Note: Once $DY$ is found, there is no need to do the trig. Notice that the hexagon consists of two trapezoids, $ABPQ$ and $CDPQ$ . $PQ = \frac{19-7-5 +11}{2} = 9$ . The height is one half of $BY$ which is $\frac{5\sqrt{3}}{4}$ . So the area is $\frac{1}{2} \cdot \frac{5\sqrt{3}}{4}(19+9+11+9)=30\sqrt{3}$

Solution 2

Let $AP$ and $BQ$ meet $CD$ at $X$ and $Y$ , respectively.

Since $\angle APD=90^{\circ}$ , $\angle ADP=\angle XDP$ , and they share $DP$ , triangles $APD$ and $XPD$ are congruent.

By the same reasoning, we also have that triangles $BQC$ and $YQC$ are congruent.

Hence, we have $[ABQCDP]=[ABYX]+\frac{[ABCD]-[ABYX]}{2}=\frac{[ABCD]+[ABYX]}{2}$ .

If we let the height of the trapezoid be $x$ , we have $[ABQCDP]=\frac{\frac{11+19}{2}\cdot x+\frac{11+7}{2}\cdot x}{2}=12x$ .

Thusly, if we find the height of the trapezoid and multiply it by 12, we will be done.

Let the projections of $A$ and $B$ to $CD$ be $A'$ and $B'$ , respectively.

We have $DA'+CB'=19-11=8$ , $DA'=\sqrt{DA^2-AA'^2}=\sqrt{49-x^2}$ , and $CB'=\sqrt{CB^2-BB'^2}=\sqrt{25-x^2}$ .

Therefore, $\sqrt{49-x^2}+\sqrt{25-x^2}=8$ . Solving this, we easily get that $x=\frac{5\sqrt{3}}{2}$ .

Multiplying this by 12, we find that the area of hexagon $ABQCDP$ is $30\sqrt{3}$ , which corresponds to answer choice $\boxed{B}$ .

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]$

$P$ is the intersection of the angle bisectors of $\angle A$ and $\angle D$ . By definition, angle bisectors are always equidistant from the sides of the angle, so $P$ is equidistant from $\overline{AB}$ , $\overline{AD}$ , and $\overline{CD}$ . Likewise, point $Q$ is equidistant from $\overline{AB}$ , $\overline{BC}$ , and $\overline{CD}$ . Because both points $P$ and $Q$ are equidistant from $\overline{AB}$ and $\overline{CD}$ and the distance between $\overline{AB}$ and $\overline{CD}$ is constant, the common distances from each of the points to the mentioned segments is equal for $P$ and $Q$ . Call this distance $x$ .

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 $P$ to $\overline{AD}$ is $x$ , so the area of $\triangle ADP$ is equal to $\frac12\cdot AD\cdot x=\frac72x$ . Similarly, the area of $\triangle BCQ$ is $\frac12\cdot BC\cdot x=\frac52x$ . The altitude of the trapezoid is $2x$ , because it is the sum of the distances from either $P$ or $Q$ to $\overline{AB}$ and $\overline{CD}$ . This means the area of trapezoid $ABCD$ is $\frac12(AB+CD)\cdot2x=\frac12(11+19)\cdot2x=30x$ . Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$ , minus the areas of triangles $ADP$ and $BCQ$ . This is $30x-\frac72x-\frac52x=24x$ . Now it remains to find $x$ .

$[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label("$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 $R$ and $S$ be the feet of the altitudes of $A$ and $B$ , respectively, to $\overline{CD}$ . We define $r=RD$ and $s=SC$ . We know that $AB=RS$ , so $RS=11$ and $r+s=19-11=8$ . By the Pythagorean Theorem on $\triangle ADR$ and $\triangle BCS$ , we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$ , respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$ . The left hand side of this equation is a difference of squares and factors to $(r+s)(r-s)$ . We know that $r+s=8$ , so $r-s=\frac{24}8=3$ . Now we can solve for $r$ by adding the two equations we just got to see that $2r=11$ , or $r=\frac{11}2$ .

We now solve for $x$ . We know that $r^2+(2x)^2=49$ , so $(2x)^2=49-\left(\frac{11}2\right)^2=\frac{75}4$ and $2x=\frac{5\sqrt3}2$ . We multiply both sides of this equation by $12$ to get $24x=30\sqrt3$ . However, the area of hexagon $ABQCDP$ is $24x$ , so the answer is $30\sqrt 3$ , or answer choice $\boxed{B}$ .

Solution 4

$[asy] import olympiad; unitsize(0.5cm); pair A, B, C, D; A = 5*(Cos(120), Sin(120)); B = A + (-11, 0); C = origin + (-19, 0); D = origin; label("$A$", A, dir(30)); label("$B$", B, dir(150)); label("$C$", C, dir(150)); label("$D$", D, dir(30)); pair E, F, G, H; E = bisectorpoint(B, A, D); F = bisectorpoint(A, B, C); G = bisectorpoint(B, C, D); H = bisectorpoint(C, D, A); pair P, Q; P = extension(A, E, D, H); Q = extension(B, F, C, G); dot("$P$", P, dir(20)); dot("$Q$", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label("$W$", W, dir(100)); label("$X$", X, dir(60)); label("$Y$", Y, dir(50)); label("$Z$", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$

Let $W, X, Y, Z = \overline{AP} \cap \overline{CD}, \overline{BQ} \cap \overline{CD}, \overline{CQ} \cap \overline{AB}, \overline{DP} \cap \overline{AB}$ respectively. Since $\angle{DCQ} = \angle{BCQ}, \angle{CBQ} = \angle{ABQ}$ we have $\angle{QCB} + \angle{CBQ} = 90 \iff \overline{BX} \perp \overline{CY};$ similarly we get $\overline{AW} \perp \overline{DZ}.$ Thus, $\overline{BQ}$ is both an angle bisector and altitude of $\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\triangle{BCX}$ gives $BC = BX \iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \frac{1}{4}\left([BYXC] + [ADWZ]\right) = 15h - \frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\sqrt{10(5)(3)(2)} = 10 \sqrt{3} = 4h \iff h = \frac{5\sqrt{3}}{2}.$ Multiply to get our answer is $\boxed{\mathrm{B}}.$

Solution 5

Like above solutions, find out the height of $ABCD$ is $\frac{5 \sqrt{3}}{2}.$ Let $M$ be the intersection of the line through $Q$ and parallel to $AB,$ and $N$ the intersection of the line through $P$ and parallel to $AB.$ Angle chasing shows that $M$ is the midpoint of $BC$ and $N$ midpoint of $AD.$ Then from midline theorem, $M, Q, N$ are collinear, and likewise for $N, P, M.$ Thus, the line through $PQ$ is in fact the midline of $ABCD.$

Let $BQ \cap CD = X, AP \cap CD = Y.$ Then, angle chasing shows that $CQ$ not only bisects $BX,$ but is also perpendicular to it. This makes it a perpendicular bisector. The same is true for $DP$ and $AY.$ Thus, $CX = CB = 5,$ and $DY = DA = 7.$ This means $XY = 19 - 5 - 7 = 7.$ We can now find $PQ$ as the midline of $ABXY.$ Thus, $PQ = \frac{1}{2} (11+7) = 9.$

Now, the answer is simply finding the area of $ABQP$ plus area of $CDPQ.$ This is $\frac{1}{2} \cdot \frac{5 \sqrt{3}}{2} \cdot \frac{11+9}{2} + \frac{1}{2} \cdot \frac{5 \sqrt{3}}{2} \cdot \frac{19+9}{2} = \boxed{30 \sqrt{3}}.$

~ MelonGirl

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Question
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