Difference between revisions of "1964 AHSME Problems/Problem 39"
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In <math>\triangle ABC</math>, since <math>c \le b \le a</math>, we have <math>\angle C \le \angle B \le \angle A</math> by the above argument. | In <math>\triangle ABC</math>, since <math>c \le b \le a</math>, we have <math>\angle C \le \angle B \le \angle A</math> by the above argument. | ||
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− | + | Now, <math>\angle AXC > \angle B \ge \angle C</math>, hence we have <math>AC > AX \implies b > AX</math> | |
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+ | And, <math>\angle BYC > \angle A \ge \angle C</math>, hence we have <math>BY > BC \implies a > BY</math> | ||
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+ | And, <math>\angle CZB > \angle A \ge \angle B</math>, hence we have <math>BC > ZC \implies a > ZC</math> | ||
Finally, adding all three inequalities, we have <math>b + a + a > AX + BY + ZC \implies AX + BY + CZ < 2a + b</math> | Finally, adding all three inequalities, we have <math>b + a + a > AX + BY + ZC \implies AX + BY + CZ < 2a + b</math> |
Revision as of 01:55, 18 September 2021
Problem
The magnitudes of the sides of triangle are
,
, and
, as shown, with
. Through interior point
and the vertices
,
,
, lines are drawn meeting the opposite sides in
,
,
, respectively. Let
. Then, for all positions of point
,
is less than:
Solution
We know that in a , if
then
, we can use this fact in the different triangles to form inequalities, and then add the inequalities.
In , since
, we have
by the above argument.
Now, , hence we have
And, , hence we have
And, , hence we have
Finally, adding all three inequalities, we have
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 38 |
Followed by Problem 40 | |
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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