Difference between revisions of "1967 AHSME Problems/Problem 32"
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+ | ==Problem== | ||
+ | |||
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> | ||
+ | |||
+ | ==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> | ||
+ | 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> | ||
+ | - 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 21:10, 29 May 2021
Contents
[hide]Problem
In quadrilateral with diagonals and , intersecting at , , , , , and . The length of is:
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 , but we want to find the value of AD. We can apply stewart's theorem now, letting , and we have , and we see that ,
Solution 2
Since , is cyclic through power of a point. From the given information, we see that and . Hence, we can find and . Letting be , we can use Ptolemy's to get Since we are solving for - 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.