Difference between revisions of "1954 AHSME Problems/Problem 29"
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==Solution== | ==Solution== | ||
| + | Let <math>\triangle ABC</math> be a right triangle with right angle at <math>B</math>, <math>AB = 2</math>, and <math>BC = 1</math>. Let the altitude from <math>B</math> to <math>AC</math> intersect <math>AC</math> at <math>D</math>. From the Pythagorean Theorem, <math>AC = \sqrt{1^2 + 2^2} = \sqrt5</math>, so <math>BD = \frac{1\cdot 2}{\sqrt5} = \frac{2\sqrt5}{5}</math>. Since <math>\triangle ADB \sim \triangle BDC \sim \triangle ABC</math>, <math>\frac{AD}{BD} = 2</math> and <math>\frac{CD}{BD} = \frac12</math>. Therefore, <math>AD = \frac{4\sqrt5}{5}</math> and <math>CD = \frac{\sqrt5}{5}</math>, so <math>CD:AD = \boxed{\textbf{(A) }1:4}</math>. | ||
==See Also== | ==See Also== | ||
Latest revision as of 00:44, 28 February 2020
Problem
If the ratio of the legs of a right triangle is 
, then the ratio of the corresponding segments of
the hypotenuse made by a perpendicular upon it from the vertex is:
Solution
Let 
be a right triangle with right angle at 
, 
, and 
. Let the altitude from to 
intersect at 
. From the Pythagorean Theorem, 
, so 
. Since 
, 
and 
. Therefore, 
and 
, so 
.
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 28 |
Followed by Problem 30 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
