Difference between revisions of "2010 USAMO Problems/Problem 4"
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<math>CE</math> meet at <math>I</math>. Determine whether or not it is possible for | <math>CE</math> meet at <math>I</math>. Determine whether or not it is possible for | ||
segments <math>AB, AC, BI, ID, CI, IE</math> to all have integer lengths. | segments <math>AB, AC, BI, ID, CI, IE</math> to all have integer lengths. | ||
+ | |||
+ | ==Video Solution in 3 minutes!!!== | ||
+ | https://www.youtube.com/watch?v=Z9fjesTOs1Q | ||
==Solution== | ==Solution== | ||
− | We know that angle <math>BIC = 135^{\circ}</math>, as the other two angles in triangle <math>BIC</math> add to <math>45^{\circ}</math>. Assume that only <math>AB, | + | We know that angle <math>BIC = 135^{\circ}</math>, as the other two angles in triangle <math>BIC</math> add to <math>45^{\circ}</math>. Assume that only <math>AB, AC, BI</math>, and <math>CI</math> are integers. Using the [[Law of Cosines]] on triangle BIC, |
<center> | <center> | ||
<asy> | <asy> | ||
Line 43: | Line 46: | ||
</center> | </center> | ||
− | <math>BC^2 = BI^2 + CI^2 - 2BI\cdot CI \cdot \cos 135^{\circ}</math>. Observing that <math>BC^2 = AB^2 + AC^2</math> and that <math>\cos 135^{\circ} = -\frac{\sqrt{2}}{2},</math> we have | + | <math>BC^2 = BI^2 + CI^2 - 2BI\cdot CI \cdot \cos 135^{\circ}</math>. Observing that <math>BC^2 = AB^2 + AC^2</math> is an integer and that <math>\cos 135^{\circ} = -\frac{\sqrt{2}}{2},</math> we have |
<center> | <center> | ||
<cmath> | <cmath> | ||
− | + | BC^2 - BI^2 - CI^2 = BI\cdot CI\cdot \sqrt{2} | |
</cmath> | </cmath> | ||
</center> | </center> | ||
Line 52: | Line 55: | ||
<center> | <center> | ||
<cmath> | <cmath> | ||
− | \sqrt{2} = \frac{ | + | \sqrt{2} = \frac{BC^2 - BI^2 - CI^2}{BI\cdot CI} |
</cmath> | </cmath> | ||
</center> | </center> | ||
− | + | The LHS (<math>\sqrt{2}</math>) is irrational, while the RHS is the quotient of the division of two integers and thus is rational. Clearly, there is a contradiction. Therefore, it is impossible for <math>AB, AC, BI</math>, and <math>CI</math> to all be integers, which invalidates the original claim that all six lengths are integers, and we are done. | |
==Solution 2== | ==Solution 2== | ||
+ | The answer is no. | ||
+ | |||
+ | Suppose otherwise. It is easy to see (through simple angle chasing) that <math>\angle DIC=45^{\circ}</math>. Also, since <math>I</math> is the incenter, we have <math>\angle IAC = 45^{\circ}</math>. Using the Law of Cosines, we have <cmath>CD^2=IC^2+ID^2-\sqrt{2}(IC)(ID),</cmath> so that <math>CD</math> is irrational. But <math>\triangle IAC \sim \triangle DIC</math>, thus <math>IC^2=CD\cdot AC</math>, implying that <math>CD</math> is rational, contradiction. <math>\blacksquare</math> | ||
+ | |||
+ | ==Solution 3== | ||
The result can be also proved without direct appeal to trigonometry, | The result can be also proved without direct appeal to trigonometry, | ||
via just the angle bisector theorem and the structure of Pythagorean | via just the angle bisector theorem and the structure of Pythagorean | ||
Line 138: | Line 146: | ||
their ratio, which is <math>2</math>, would be the square of a rational number, | their ratio, which is <math>2</math>, would be the square of a rational number, | ||
but <math>\sqrt{2}</math> is irrational, and so the assumed triangle cannot exist. | but <math>\sqrt{2}</math> is irrational, and so the assumed triangle cannot exist. | ||
+ | |||
+ | ==Solution 4== | ||
+ | We proceed by contradiction. | ||
+ | |||
+ | FTSOC, let <math>AB, BI, ID</math> have integer lengths. Then <math>BD = BI + ID \in \mathbb{Z}</math> as well. By trigonometry, <cmath>BD = \frac{AB}{\cos{(\frac{\angle ABC}{2})}}.</cmath> | ||
+ | Rearranging we find <math>\cos{(\frac{\angle ABC}{2})} = \frac{AB}{BD} \in \mathbb{Q}</math>. As <math>0 < \angle ABC < \frac{\pi}{2}</math> so <math>0 < \frac{\angle ABC}{2} < \frac{\pi}{4}</math> but there are no possible angles in the interval that result in a rational cosine by Niven's theorem. So we have contradiction. | ||
+ | |||
+ | ~Aaryabhatta1. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=Dh9H_DNDMAg | ||
==See also== | ==See also== | ||
{{USAMO newbox|year=2010|num-b=3|num-a=5}} | {{USAMO newbox|year=2010|num-b=3|num-a=5}} | ||
+ | {{USAJMO newbox|year=2010|num-b=5|after=Last Problem}} | ||
+ | |||
+ | [[Category: Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:47, 31 January 2024
Contents
Problem
Let be a triangle with
. Points
and
lie on sides
and
, respectively, such that
and
. Segments
and
meet at
. Determine whether or not it is possible for
segments
to all have integer lengths.
Video Solution in 3 minutes!!!
https://www.youtube.com/watch?v=Z9fjesTOs1Q
Solution
We know that angle , as the other two angles in triangle
add to
. Assume that only
, and
are integers. Using the Law of Cosines on triangle BIC,
. Observing that
is an integer and that
we have
and therefore,
The LHS () is irrational, while the RHS is the quotient of the division of two integers and thus is rational. Clearly, there is a contradiction. Therefore, it is impossible for
, and
to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.
Solution 2
The answer is no.
Suppose otherwise. It is easy to see (through simple angle chasing) that . Also, since
is the incenter, we have
. Using the Law of Cosines, we have
so that
is irrational. But
, thus
, implying that
is rational, contradiction.
Solution 3
The result can be also proved without direct appeal to trigonometry, via just the angle bisector theorem and the structure of Pythagorean triples. (This is a lot more work).
A triangle in which all the required lengths are integers exists if and
only if there exists a triangle in which and
are
relatively-prime integers and the lengths of the segments
are all rational
(we divide all the lengths by the
or
conversely multiply all the lengths by the least common multiple
of the denominators of the rational lengths).
Suppose there exists a triangle in which the lengths and
are
relatively-prime integers and the lengths
are all rational.
Since is the bisector of
, by the angle bisector
theorem, the ratio
, and since
is the
bisector of
,
. Therefore,
. Now
is by assumption rational,
so
is rational, but
and
are assumed integers
so
must also be rational. Since
is the hypotenuse of
a right-triangle, its length is the square root of an integer,
and thus either an integer or irrational, so
must be an integer.
With and
relatively-prime, we conclude that the side
lengths of
must be a Pythagorean triple:
, with
relatively-prime positive integers
and
odd.
Without loss of generality, .
By the angle bisector theorem,
Since is a right-triangle, we have:
and so is rational if and only if
is a perfect square.
Also by the angle bisector theorem,
and therefore, since is a right-triangle, we have:
and so is rational if and only if
is a perfect square.
Combining the conditions on and
, we see that
and
must both be perfect squares. If it were so,
their ratio, which is
, would be the square of a rational number,
but
is irrational, and so the assumed triangle cannot exist.
Solution 4
We proceed by contradiction.
FTSOC, let have integer lengths. Then
as well. By trigonometry,
Rearranging we find
. As
so
but there are no possible angles in the interval that result in a rational cosine by Niven's theorem. So we have contradiction.
~Aaryabhatta1.
Video Solution
https://www.youtube.com/watch?v=Dh9H_DNDMAg
See also
2010 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
2010 USAJMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.