Difference between revisions of "1987 OIM Problems/Problem 2"
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== Problem == | == Problem == | ||
| − | On a triangle <math>ABC</math>, <math>M</math> and <math>N</math> are the respective midpoints of sides <math>AC</math> and <math>AB</math>, and <math>P</math> is the | + | On a triangle <math>ABC</math>, <math>M</math> and <math>N</math> are the respective midpoints of sides <math>AC</math> and <math>AB</math>, and <math>P</math> is the centroid of <math>\triangle ABC</math>. Prove that, if is possible to inscribe a circle in the quadrilateral <math>ANPM</math>, then triangle <math>ABC</math> is isosceles. |
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ||
== Solution == | == Solution == | ||
| − | {{ | + | By the [[Pitot Theorem]], it is possible to inscribe such a circle if and only if <math>AN+MP=PN+MA</math>. As a result, we can substitute. We know that <math>AN=\frac{1}{2}c</math> and <math>MA=\frac{1}{2}b</math>. Furthermore, the formula for the length of the median from <math>A</math> of any triangle <math>\triangle ABC</math> is |
| + | <math><cmath>\frac{1}{2}\sqrt{2b^2+2c^2-a^2}</cmath> | ||
| + | Using this and the </math>2:1<math> ratio property of the median, we must have | ||
| + | <cmath>PN=\frac{1}{6}\sqrt{2a^2+2b^2-c^2}</cmath> | ||
| + | <cmath>MP=\frac{1}{6}\sqrt{2a^2+2c^2-b^2}</cmath> | ||
| + | Combining the above, we find that | ||
| + | <cmath>\frac{1}{2}c+\frac{1}{6}\sqrt{2a^2+2c^2-b^2}=\frac{1}{6}\sqrt{2a^2+2b^2-c^2}+\frac{1}{2}b</cmath> | ||
| + | which is equivalent to | ||
| + | <cmath>\sqrt{2a^2+2c^2-b^2}-\sqrt{2a^2+2b^2-c^2}=3(b-c)</cmath> | ||
| + | Let </math>x=\sqrt{2a^2+2c^2-b^2}<math> and </math>y=\sqrt{2a^2+2b^2-c^2}<math> for simplicity. Then </math>x-y=3(b-c)<math>. If </math>b>c<math>, then analyzing the square roots yields </math>x<y<math>, but the right-hand side is positive, a contradiction. Similarly, if </math>c>b<math>, then </math>x>y<math>, but the right-hand side is negative, also a contradiction. Thus we must have </math>b=c$, and testing this case shows that it does indeed work, finishing the proof. | ||
| + | |||
| + | ~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406] | ||
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe2.htm | https://www.oma.org.ar/enunciados/ibe2.htm | ||
Revision as of 17:47, 22 March 2025
Problem
On a triangle 
, 
and 
are the respective midpoints of sides 
and 
, and 
is the centroid of 
. Prove that, if is possible to inscribe a circle in the quadrilateral 
, then triangle is isosceles.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
By the Pitot Theorem, it is possible to inscribe such a circle if and only if 
. As a result, we can substitute. We know that 
and 
. Furthermore, the formula for the length of the median from 
of any triangle is
2:1
x=\sqrt{2a^2+2c^2-b^2}
y=\sqrt{2a^2+2b^2-c^2}
x-y=3(b-c)
b>c
x<y
c>b
x>y
b=c$, and testing this case shows that it does indeed work, finishing the proof.
