Difference between revisions of "1998 JBMO Problems/Problem 2"
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+ | ==Problem 2== | ||
+ | |||
+ | Let <math>ABCDE</math> be a convex pentagon such that <math>AB=AE=CD=1</math>, <math>\angle ABC=\angle DEA=90^\circ</math> and <math>BC+DE=1</math>. Compute the area of the pentagon. | ||
+ | |||
+ | |||
+ | == Solutions == | ||
+ | |||
+ | === Solution 1 === | ||
+ | |||
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 === | ||
+ | |||
+ | Let <math>BC = x, DE = y</math>. Denote the area of <math>\triangle XYZ</math> by <math>[XYZ]</math>. | ||
+ | |||
+ | <math>[ABC]+[AED]=\frac{1}{2}(x+y)=\frac{1}{2}</math> | ||
+ | |||
+ | <math>[ACD]</math> can be found by [[Heron's formula]]. | ||
+ | |||
+ | <math>AC=\sqrt{x^2+1}</math> | ||
+ | |||
+ | <math>AD=\sqrt{y^2+1}</math> | ||
+ | |||
+ | Let <math>AC=b, AD=c</math>. | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | [ACD]&=\frac{1}{4}\sqrt{(1+b+c)(-1+b+c)(1-b+c)(1+b-c)}\\ | ||
+ | &=\frac{1}{4}\sqrt{((b+c)^2-1)(1-(b-c)^2)}\\ | ||
+ | &=\frac{1}{4}\sqrt{(b+c)^2+(b-c)^2-(b^2-c^2)^2-1}\\ | ||
+ | &=\frac{1}{4}\sqrt{2(b^2+c^2)-(b^2-c^2)^2-1}\\ | ||
+ | &=\frac{1}{4}\sqrt{2(x^2+y^2+2)-(x^2-y^2)^2-1}\\ | ||
+ | &=\frac{1}{4}\sqrt{2((x+y)^2-2xy+2)-(x+y)^2(x-y)^2-1}\\ | ||
+ | &=\frac{1}{4}\sqrt{5-4xy-(x-y)^2}\\ | ||
+ | &=\frac{1}{4}\sqrt{5-(x+y)^2}\\ | ||
+ | &=\frac{1}{2} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Total area <math>=[ABC]+[AED]+[ACD]=\frac{1}{2}+\frac{1}{2}=1</math>. | ||
+ | |||
+ | By durianice | ||
+ | |||
+ | === Solution 3 === | ||
+ | Construct <math>AD</math> and <math>AC</math> to partition the figure into <math>ABC</math>, <math>ACD</math> and <math>ADE</math>. | ||
+ | |||
+ | Rotate <math>ADE</math> with centre <math>A</math> such that <math>AE</math> coincides with <math>AB</math> and <math>AD</math> is mapped to <math>AD'</math>. Hence the area of the pentagon is still preserved and it suffices to find the area of the quadrilateral <math>AD'CD</math>. | ||
+ | |||
+ | Hence <math>[AD'C]</math> = <math>\frac{1}{2}</math> (<math>D'E + BC</math>)<math>AB</math>= <math>\frac{1}{2}</math> | ||
+ | |||
+ | Since <math>CD</math> = <math>CD'</math>, <math>AC</math> = <math>AC</math> and <math>AD</math> = <math>AD'</math>, by SSS Congruence, <math>ACD</math> and <math>ACD'</math> are congruent, so <math>[ACD]</math> = <math>\frac{1}{2}</math> | ||
+ | |||
+ | So the area of pentagon <math>ABCDE = \frac{1}{2} + \frac{1}{2} = 1</math>. | ||
+ | |||
+ | - SomebodyYouUsedToKnow | ||
+ | |||
+ | ==See Also== | ||
+ | {{JBMO box|year=1998|num-b=1|num-a=3|five=}} |
Latest revision as of 07:31, 2 July 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
Solution 3
Construct and to partition the figure into , and .
Rotate with centre such that coincides with and is mapped to . Hence the area of the pentagon is still preserved and it suffices to find the area of the quadrilateral .
Hence = ()=
Since = , = and = , by SSS Congruence, and are congruent, so =
So the area of pentagon .
- SomebodyYouUsedToKnow
See Also
1998 JBMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |