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

(→‎Solution 3 (Law of Cosines Cheese))
(→‎Solution 3 (Law of Cosines Cheese))
 
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==Solution 3 (Law of Cosines Cheese)==
 
==Solution 3 (Law of Cosines Cheese)==
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The solution says it all. Since <math>\angle AOD</math> is supplementary to <math>\angle AOB</math>, <math>cos(\angle AOD) = cos(180 \degree - \angle AOB)=-cos(\angle AOB) </math>. The law of cosines on <math>\triangle AOB</math> gives us <math>cos(\angle AOB)=\frac {8^2+4^2-6^2}{(2)(8)(4)}=\frac {11}{16}</math>. Again, we can use the law of cosines on <math>\triangle AOD</math>, which gives us  
+
The solution says it all. Since <math>\angle AOD</math> is supplementary to <math>\angle AOB</math>, <math>cos(\angle AOD) = cos(180^{\circ} - \angle AOB)=-cos(\angle AOB) </math>. The law of cosines on <math>\triangle AOB</math> gives us <math>cos(\angle AOB)=\frac {8^2+4^2-6^2}{(2)(8)(4)}=\frac {11}{16}</math>. Again, we can use the law of cosines on <math>\triangle AOD</math>, which gives us  
 
<cmath>AD=\sqrt {8^2+6^2-2(8)(6)cos(\angle AOD)}</cmath>
 
<cmath>AD=\sqrt {8^2+6^2-2(8)(6)cos(\angle AOD)}</cmath>
−
<cmath>=\sqrt {8^2+6^2-(2)(8)(6)cos(\angle 180 \degree -AOB)}</cmath>
+
<cmath>=\sqrt {8^2+6^2-(2)(8)(6)cos(\angle 180^{\circ} - AOB)}</cmath>
 
<cmath>=\sqrt {8^2+6^2+(2)(8)(6)cos(\angle AOB)}</cmath>
 
<cmath>=\sqrt {8^2+6^2+(2)(8)(6)cos(\angle AOB)}</cmath>
 
<cmath>=\sqrt {8^2+6^2+(2)(8)(6)(\frac {11}{16})}</cmath>
 
<cmath>=\sqrt {8^2+6^2+(2)(8)(6)(\frac {11}{16})}</cmath>

Latest revision as of 17:56, 17 December 2023

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: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$

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 $x = \boxed{\textbf{(E)}~\sqrt{166}}$

Solution 2

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5cm); 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("$B$", (-4.42,4), NE * labelscalefactor); dot((-3.1,-3.42),dotstyle); label("$C$", (-3.4,-3.94), NE * labelscalefactor); dot((1.56,-0.22),dotstyle); label("$D$", (1.8,-0.4), 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] (Diagram not to scale)

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

Solution 3 (Law of Cosines Cheese)

The solution says it all. Since $\angle AOD$ is supplementary to $\angle AOB$, $cos(\angle AOD) = cos(180^{\circ} - \angle AOB)=-cos(\angle AOB)$. The law of cosines on $\triangle AOB$ gives us $cos(\angle AOB)=\frac {8^2+4^2-6^2}{(2)(8)(4)}=\frac {11}{16}$. Again, we can use the law of cosines on $\triangle AOD$, which gives us \[AD=\sqrt {8^2+6^2-2(8)(6)cos(\angle AOD)}\] \[=\sqrt {8^2+6^2-(2)(8)(6)cos(\angle 180^{\circ} - AOB)}\] \[=\sqrt {8^2+6^2+(2)(8)(6)cos(\angle AOB)}\] \[=\sqrt {8^2+6^2+(2)(8)(6)(\frac {11}{16})}\] \[=\sqrt{166}\] which gives us $\boxed{\textbf{(E)}~\sqrt{166}}$

Note that this solution works even if the quadrilateral is not cyclic, and in general, it works if an angle's supplement is known. (Anyone come from aops volume 2 lmao.)

-Wesssslili

See also

1967 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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