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Difference between revisions of "1970 Canadian MO Problems/Problem 2"

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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 $ABC$ with angle $A$ obtuse and with altitudes of length $h$ and $k$ as shown in the diagram, prove that $a+h\ge b+k$. Find under what conditions $a+h=b+k$.


[asy] draw((0,0)--(5,0)--(16/5,12/5)--cycle,dot); draw((2.5,0)--(2.5,7.5/4)--(5,0)--cycle,black); MP("C",(0,0),SW);MP("D",(16/5,12/5),N);MP("B",(5,0),SE); MP("E",(2.5,0),NE);MP("A",(2.5,7.5/4),N); MP("h",(2.5,7.5/8),W);MP("k",(41/10,6/5),NE); draw((-.2,.2)--(2.5-.2,7.5/4+.2),arrow=ArcArrow(TeXHead)); draw((2.5-.2,7.5/4+.2)--(-.2,.2),arrow=ArcArrow(TeXHead)); MP("b",(2.3/2-.05,7.5/8+.25),N); draw((0,-.2)--(5,-.2),arrow=ArcArrow(TeXHead)); draw((5,-.2)--(0,-.2),arrow=ArcArrow(TeXHead)); MP("a",(2.5,-.2),S); draw((16/5,12/5)--(16/5-.2,12/5-.15)--(16/5-.2+.15,12/5-.15-.2)--(16/5+.15,12/5-.2)--cycle,black); [/asy]


Solution

There is, in fact, no equality case: $a + h > b + k$. In triangle $ACE$, we have $b > h$ since it is a right triangle. Since angle $A$ is obtuse we have $a > b$, or $(a-b) > 0$. Then $(a-b) > h(a-b)/b$, or $a - b > ha/b - h$. Here we can use the fact that $(a,h)$ and $(b,k)$ are base-altitude pairs so $k = ha/b$. Therefore $a - b > k - h$, so $a + h > b + k$.

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