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

(→‎Solution 1)
(→‎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 - \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 <math>AD = \sqrt {8^2+6^2-2(8)(6)cos(\angle AOD)} = \sqrt {8^2+6^2+2(8)(6)cos(\angle 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}</math>, which gives us <math>\boxed{E \sqrt{166}}</math>.
+
The solution says it all. Since <math>\angle AOD</math> is supplementary to <math>\angle AOB</math>, <math>cos(\angle AOD) = cos(180 - \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 <math>AD=\sqrt {8^2+6^2-2(8)(6)cos(\angle AOD)}=\sqrt {8^2+6^2+(2)(8)(6)cos(\angle 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}</math>, which gives us <math>\boxed{E \sqrt{166}}</math>.
  
 
Note that this solution works even if the quadrilateral is not cyclic, and in general, it works if an angle's supplement is known. So imo, it's better jk.
 
Note that this solution works even if the quadrilateral is not cyclic, and in general, it works if an angle's supplement is known. So imo, it's better jk.

Revision as of 17:51, 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 $\angle AOB$, $cos(\angle AOD) = cos(180 - \angle AOB)=-cos(\angle AOB)$. 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 $AD=\sqrt {8^2+6^2-2(8)(6)cos(\angle AOD)}=\sqrt {8^2+6^2+(2)(8)(6)cos(\angle AOB)}=\sqrt {8^2+6^2+(2)(8)(6)cos(\angle AOB)}=\sqrt {8^2+6^2+(2)(8)(6)(\frac {11}{16})}=\sqrt{166}$, which gives us $\boxed{E \sqrt{166}}$.

Note that this solution works even if the quadrilateral is not cyclic, and in general, it works if an angle's supplement is known. So imo, it's better jk.

-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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