Difference between revisions of "1967 AHSME Problems/Problem 32"
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==Solution 3 (Law of Cosines Cheese)== | ==Solution 3 (Law of Cosines Cheese)== | ||
− | 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 | + | 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^{\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> | + | <cmath>=\sqrt {8^2+6^2+(2)(8)(6)(\frac {11}{16})}</cmath> |
<cmath>=\sqrt{166}</cmath> | <cmath>=\sqrt{166}</cmath> | ||
which gives us <math>\boxed{\textbf{(E)}~\sqrt{166}}</math> | which gives us <math>\boxed{\textbf{(E)}~\sqrt{166}}</math> |
Latest revision as of 18:56, 17 December 2023
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
. We can apply stewart's theorem now, letting
, and we have
, and we see that
Solution 2
(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
Since we are solving for
- PhunsukhWangdu
Solution 3 (Law of Cosines Cheese)
The solution says it all. Since is supplementary to
,
. The law of cosines on
gives us
. Again, we can use the law of cosines on
, which gives us
which gives us
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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