1970 AHSME Problems/Problem 28: Difference between revisions
Created page with "== Problem == In triangle <math>ABC</math>, the median from vertex <math>A</math> is perpendicular to the median from vertex <math>B</math>. If the lengths of sides <math>AC</ma..." |
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== Solution == | == Solution == | ||
<math>\fbox{A}</math> | <math>\fbox{A}</math> | ||
let the midpoint be M,N ( i.e. AM,BN are the medians); connecting MN we know that AB = 2x and MN = x hence apply stewart's theorem in triangle ABC with median MN first and then apply stewart's in triangle BNC with median MN | |||
== See also == | == See also == | ||
Latest revision as of 01:36, 19 December 2020
Problem
In triangle 
, the median from vertex 
is perpendicular to the median from vertex 
. If the lengths of sides 
and 
are 
and 
respectively, then the length of side 
is
Solution
let the midpoint be M,N ( i.e. AM,BN are the medians); connecting MN we know that AB = 2x and MN = x hence apply stewart's theorem in triangle ABC with median MN first and then apply stewart's in triangle BNC with median MN
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 27 |
Followed by Problem 29 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
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