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Difference between revisions of "1967 AHSME Problems/Problem 32"

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==Problem==
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In quadrilateral <math>ABCD</math> with diagonals <math>AC</math> and <math>BD</math>, intersecting at <math>O</math>, <math>BO=4</math>, <math>OD = 6</math>, <math>AO=8</math>, <math>OC=3</math>, and <math>AB=6</math>. The length of <math>AD</math> is:
 
In quadrilateral <math>ABCD</math> with diagonals <math>AC</math> and <math>BD</math>, intersecting at <math>O</math>, <math>BO=4</math>, <math>OD = 6</math>, <math>AO=8</math>, <math>OC=3</math>, and <math>AB=6</math>. The length of <math>AD</math> is:
 
<math>\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}</math>
 
<math>\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}</math>
  
 
<math>\sqrt{166}</math>
 
<math>\sqrt{166}</math>
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 +
==Solution 1==
  
 
After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of <math>AB, BD(BD = BO + OD)</math>, but we want to find the value of AD. We can apply stewart's theorem now, letting <math>m = 4, n = 6, AD = X, AB = 6</math>, and we have <math>10 \cdot 6 \cdot 4 + 8 \cdot 8 \cdot 10 = x^2 + 36 \cdot 6</math>, and we see that <math>x = \sqrt{166}</math>, <math>\boxed{E \sqrt{166}}</math>
 
After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of <math>AB, BD(BD = BO + OD)</math>, but we want to find the value of AD. We can apply stewart's theorem now, letting <math>m = 4, n = 6, AD = X, AB = 6</math>, and we have <math>10 \cdot 6 \cdot 4 + 8 \cdot 8 \cdot 10 = x^2 + 36 \cdot 6</math>, and we see that <math>x = \sqrt{166}</math>, <math>\boxed{E \sqrt{166}}</math>
 +
 +
==Solution 2==
 +
 +
<asy>
 +
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
 +
import graph; size(2cm);
 +
real labelscalefactor = 0.5; /* changes label-to-point distance */
 +
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
 +
pen dotstyle = black; /* point style */
 +
real xmin = -9.78, xmax = 9.78, ymin = -5.72, ymax = 5.72;  /* image dimensions */
 +
 +
 +
draw((-1,4)--(-4.08,3.78)--(-3.1,-3.42)--(1.56,-0.22)--cycle, linewidth(2));
 +
/* draw figures */
 +
draw((-1,4)--(-4.08,3.78), linewidth(2));
 +
draw((-4.08,3.78)--(-3.1,-3.42), linewidth(2));
 +
draw((-3.1,-3.42)--(1.56,-0.22), linewidth(2));
 +
draw((1.56,-0.22)--(-1,4), linewidth(2));
 +
draw(circle((-2.287661623108666,0.35726272352132055), 3.863626188061437), linewidth(2));
 +
draw((-1,4)--(-3.1,-3.42), linewidth(2));
 +
draw((-4.08,3.78)--(1.56,-0.22), linewidth(2));
 +
/* dots and labels */
 +
dot((-1,4),dotstyle);
 +
label("$A$", (-0.92,4.2), NE * labelscalefactor);
 +
dot((-4.08,3.78),dotstyle);
 +
label("$B$", (-4,3.98), NE * labelscalefactor);
 +
dot((-3.1,-3.42),dotstyle);
 +
label("$C$", (-3.02,-3.22), NE * labelscalefactor);
 +
dot((1.56,-0.22),dotstyle);
 +
label("$D$", (1.64,-0.02), NE * labelscalefactor);
 +
dot((-1.566733533935139,1.9975415134291763),linewidth(4pt) + dotstyle);
 +
label("$O$", (-1.34,1.98), NE * labelscalefactor);
 +
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 +
/* end of picture */
 +
</asy>
 +
Since <math>AO \cdot OC = BO \cdot OD</math>, <math>ABCD</math> is cyclic through power of a point. From the given information, we see that <math>\triangle{AOB}\sim \triangle{DOC}</math> and <math>\triangle{BOC} \sim \triangle{AOD}</math>. Hence, we can find <math>CD=\frac{9}{2}</math> and <math>AD=2 \cdot BC</math>. Letting <math>BC</math> be <math>x</math>, we can use Ptolemy's to get
 +
<cmath>6 \cdot \frac{9}{2} + 2x^2=10 \cdot 11 \implies x=\sqrt{\frac{83}{2}}</cmath>
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Since we are solving for <math>AD=2x=2\cdot\sqrt{\frac{83}{2}}=\sqrt{4\cdot\frac{83}{2}} = \boxed{\textbf{(E)}~\sqrt{166}}</math>
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- PhunsukhWangdu
 +
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1967|num-b=31|num-a=33}}   
 
{{AHSME box|year=1967|num-b=31|num-a=33}}   

Revision as of 22:10, 29 May 2021

Problem

In quadrilateral $ABCD$ with diagonals $AC$ and $BD$, intersecting at $O$, $BO=4$, $OD = 6$, $AO=8$, $OC=3$, and $AB=6$. The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$

$\sqrt{166}$

Solution 1

After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of $AB, BD(BD = BO + OD)$, but we want to find the value of AD. We can apply stewart's theorem now, letting $m = 4, n = 6, AD = X, AB = 6$, and we have $10 \cdot 6 \cdot 4 + 8 \cdot 8 \cdot 10 = x^2 + 36 \cdot 6$, and we see that $x = \sqrt{166}$, $\boxed{E \sqrt{166}}$

Solution 2

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(2cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.78, xmax = 9.78, ymin = -5.72, ymax = 5.72; /* image dimensions */ draw((-1,4)--(-4.08,3.78)--(-3.1,-3.42)--(1.56,-0.22)--cycle, linewidth(2)); /* draw figures */ draw((-1,4)--(-4.08,3.78), linewidth(2)); draw((-4.08,3.78)--(-3.1,-3.42), linewidth(2)); draw((-3.1,-3.42)--(1.56,-0.22), linewidth(2)); draw((1.56,-0.22)--(-1,4), linewidth(2)); draw(circle((-2.287661623108666,0.35726272352132055), 3.863626188061437), linewidth(2)); draw((-1,4)--(-3.1,-3.42), linewidth(2)); draw((-4.08,3.78)--(1.56,-0.22), linewidth(2)); /* dots and labels */ dot((-1,4),dotstyle); label("$A$", (-0.92,4.2), NE * labelscalefactor); dot((-4.08,3.78),dotstyle); label("$B$", (-4,3.98), NE * labelscalefactor); dot((-3.1,-3.42),dotstyle); label("$C$", (-3.02,-3.22), NE * labelscalefactor); dot((1.56,-0.22),dotstyle); label("$D$", (1.64,-0.02), NE * labelscalefactor); dot((-1.566733533935139,1.9975415134291763),linewidth(4pt) + dotstyle); label("$O$", (-1.34,1.98), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy] Since $AO \cdot OC = BO \cdot OD$, $ABCD$ is cyclic through power of a point. From the given information, we see that $\triangle{AOB}\sim \triangle{DOC}$ and $\triangle{BOC} \sim \triangle{AOD}$. Hence, we can find $CD=\frac{9}{2}$ and $AD=2 \cdot BC$. Letting $BC$ be $x$, we can use Ptolemy's to get \[6 \cdot \frac{9}{2} + 2x^2=10 \cdot 11 \implies x=\sqrt{\frac{83}{2}}\] Since we are solving for $AD=2x=2\cdot\sqrt{\frac{83}{2}}=\sqrt{4\cdot\frac{83}{2}} = \boxed{\textbf{(E)}~\sqrt{166}}$ - PhunsukhWangdu

See also

1967 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 31 		Followed by
Problem 33
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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