Difference between revisions of "1970 Canadian MO Problems/Problem 2"
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== Solution == | == Solution == | ||
+ | There is, in fact, no equality case: <math>a + h > b + k</math>. In triangle <math>ACE</math>, we have <math>b > h</math> since it is a right triangle. Since angle <math>A</math> is obtuse we have <math>a > b</math>, or <math>(a-b) > 0</math>. Then <math>(a-b) > h(a-b)/b</math>, or <math>a - b > ha/b - h</math>. Here we can use the fact that <math>(a,h)</math> and <math>(b,k)</math> are base-altitude pairs so <math>k = ha/b</math>. Therefore <math>a - b > k - h</math>, so <math>a + h > b + k</math>. |
Latest revision as of 19:32, 5 June 2018
Problem
Given a triangle with angle
obtuse and with altitudes of length
and
as shown in the diagram, prove that
. Find under what conditions
.
Solution
There is, in fact, no equality case: . In triangle
, we have
since it is a right triangle. Since angle
is obtuse we have
, or
. Then
, or
. Here we can use the fact that
and
are base-altitude pairs so
. Therefore
, so
.