Difference between revisions of "1970 IMO Problems/Problem 1"
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But this follows from the fact that the angles <math>AMC</math> and <math>CBM</math> are supplementary. | But this follows from the fact that the angles <math>AMC</math> and <math>CBM</math> are supplementary. | ||
− | + | ==Solution 2== | |
+ | By similar triangles and the fact that both centers lie on the angle bisector of <math>\angle{C}</math>, we have <math>\frac{r}{q} = \frac{s-c}{s} = \frac{a + b - c}{a + b + c}</math>, where <math>s</math> is the semi-perimeter of <math>ABC</math>. Let <math>ABC</math> have sides <math>a, b, c</math>, and let <math>AM = c_1, MB = c_2, MC = d</math>. After simple computations, we see that the condition, whose equivalent form is | ||
+ | <cmath>\frac{b + d - c_1}{b + d + c_1} \cdot \frac{a + d - c_2}{a + d + c_2} = \frac{a + b - c}{a + b + c},</cmath> | ||
+ | is also equivalent to Stewart's Inequality | ||
+ | <cmath>d^2 c + c_1 c_2 c = a^2 c_1 + b^2 c_2.</cmath> | ||
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 21:12, 19 May 2015
Problem
Let be a point on the side of . Let , and be the inscribed circles of triangles , and . Let , and be the radii of the exscribed circles of the same triangles that lie in the angle . Prove that
.
Solution
We use the conventional triangle notations.
Let be the incenter of , and let be its excenter to side . We observe that
,
and likewise,
Simplifying the quotient of these expressions, we obtain the result
.
Thus we wish to prove that
.
But this follows from the fact that the angles and are supplementary.
Solution 2
By similar triangles and the fact that both centers lie on the angle bisector of , we have , where is the semi-perimeter of . Let have sides , and let . After simple computations, we see that the condition, whose equivalent form is is also equivalent to Stewart's Inequality Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1970 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |