Difference between revisions of "1998 JBMO Problems/Problem 2"
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Let <math>BC = a, ED = 1 - a</math> | Let <math>BC = a, ED = 1 - a</math> | ||
− | Let | + | Let <math>\angle DAC = X</math> |
− | Applying cosine rule to | + | Applying cosine rule to <math>\triangle DAC</math> we get: |
− | <math> | + | <math>\cos X = \frac{AC ^ {2} + AD ^ {2} - DC ^ {2}}{ 2 \cdot AC \cdot AD }</math> |
Substituting <math>AC^{2} = 1^{2} + a^{2}, AD ^ {2} = 1^{2} + (1-a)^{2}, DC = 1</math> we get: | Substituting <math>AC^{2} = 1^{2} + a^{2}, AD ^ {2} = 1^{2} + (1-a)^{2}, DC = 1</math> we get: | ||
− | <math> | + | <math>\cos^{2} X = \frac{(1 - a - a ^ {2}) ^ {2}}{((1 + a^{2})(2 - 2a + a^{2}))}</math> |
− | From above, <math> | + | From above, <math>\sin^{2} X = 1 - \cos^{2} X = \frac{1}{((1 + a^{2})(2 - 2a + a^{2}))} = \frac{1}{AC^{2} \cdot AD^{2}}</math> |
− | Thus, <math> | + | Thus, <math>\sin X \cdot AC \cdot AD = 1</math> |
− | So, <math> | + | So, area of <math>\triangle DAC</math> = <math>\frac{1}{2}\cdot \sin X \cdot AC \cdot AD = \frac{1}{2}</math> |
− | Let <math>AF</math> be the altitude of triangle DAC from A. | + | Let <math>AF</math> be the altitude of <math>\triangle DAC</math> from <math>A</math>. |
− | So <math>1 | + | So <math>\frac{1}{2}\cdot DC\cdot AF = \frac{1}{2}</math> |
This implies <math>AF = 1</math>. | This implies <math>AF = 1</math>. | ||
− | Since <math>AFCB</math> is a cyclic quadrilateral with <math>AB = AF</math>, | + | Since <math>AFCB</math> is a cyclic quadrilateral with <math>AB = AF</math>, <math>\triangle ABC</math> is congruent to <math>\triangle AFC</math>. |
− | Similarly <math>AEDF</math> is a cyclic quadrilateral and | + | Similarly <math>AEDF</math> is a cyclic quadrilateral and <math>\triangle AED</math> is congruent to <math>\triangle AFD</math>. |
− | So | + | So area of <math>\triangle ABC</math> + area of <math>\triangle AED</math> = area of <math>\triangle ADC</math>. |
− | Thus | + | Thus area of pentagon <math>ABCD</math> = area of <math>\triangle ABC</math> + area of <math>\triangle AED</math> + area of <math>\triangle DAC</math> = <math>\frac{1}{2}+\frac{1}{2} = 1</math> |
By <math>Kris17</math> | By <math>Kris17</math> | ||
− | |||
=== Solution 2 === | === Solution 2 === |
Revision as of 22:23, 4 June 2020
Problem 2
Let be a convex pentagon such that , and . Compute the area of the pentagon.
Solutions
Solution 1
Let
Let
Applying cosine rule to we get:
Substituting we get:
From above,
Thus,
So, area of =
Let be the altitude of from .
So
This implies .
Since is a cyclic quadrilateral with , is congruent to . Similarly is a cyclic quadrilateral and is congruent to .
So area of + area of = area of . Thus area of pentagon = area of + area of + area of =
By
Solution 2
Let . Denote the area of by .
can be found by Heron's formula.
Let .
Total area .
By durianice