Difference between revisions of "1967 AHSME Problems/Problem 32"
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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 | 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> | <cmath>6 \cdot \frac{9}{2} + 2x^2=10 \cdot 11 \implies x=\sqrt{\frac{83}{2}}</cmath> | ||
Revision as of 21:12, 29 May 2021
Contents
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
![[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]](http://latex.artofproblemsolving.com/8/8/7/8871207e2536eed5f290421ae5dab393c5d0455e.png)
(Diagram not to scale)
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
![\[6 \cdot \frac{9}{2} + 2x^2=10 \cdot 11 \implies x=\sqrt{\frac{83}{2}}\]](http://latex.artofproblemsolving.com/6/c/e/6cee7a785de15773e94d64e6a5fa0d3cbf32c25a.png)
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. 
