Difference between revisions of "1991 AIME Problems/Problem 14"
m |
Someperson01 (talk | contribs) (Solution Done.) |
||
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
| − | + | Let x=AC, y=AD, and z=AE. | |
| + | Ptolemy's Theorem on ABCD gives <math>81y+31\cdot 81=xz</math>, and Ptolemy on ACDE gives <math>x\cdot z+81^2=y^2</math>. | ||
| + | Subtracting these equations give <math>y^2-81y-112\cdot 81=0</math>, and from this y=144. Ptolemy on ADEF gives <math>81y+81^2=z^2</math>, and from this z=135. Finally, plugging back into the first equation gives x=105, so x+y+z=105+144+135=<math>384</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1991|num-b=13|num-a=15}} | {{AIME box|year=1991|num-b=13|num-a=15}} | ||
Revision as of 11:41, 21 October 2007
Problem
A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted by 
, has length 31. Find the sum of the lengths of the three diagonals that can be drawn from 
.
Solution
Let x=AC, y=AD, and z=AE.
Ptolemy's Theorem on ABCD gives 
, and Ptolemy on ACDE gives 
.
Subtracting these equations give 
, and from this y=144. Ptolemy on ADEF gives 
, and from this z=135. Finally, plugging back into the first equation gives x=105, so x+y+z=105+144+135=
.
See also
| 1991 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
