Difference between revisions of "2020 USAMTS Round 1 Problems/Problem 3"
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By the same logic, <math>HF \parallel EG \Rightarrow GFHE</math> is a parallelogram. | By the same logic, <math>HF \parallel EG \Rightarrow GFHE</math> is a parallelogram. | ||
2. <math>\angle EAB = \frac{\angle DAB}{2}</math> and <math>\angle ABE = \frac{\angle ABC}{2} \Rightarrow \angle EAB + \angle ABE = \frac{\angle DAB + \angle ABC}{2}</math> and <math>\angle DAB + \angle ABC = 180^\circ \Rightarrow \angle EAB + \angle ABE = 90^\circ \Rightarrow \angle AEB = 90^\circ.</math> | 2. <math>\angle EAB = \frac{\angle DAB}{2}</math> and <math>\angle ABE = \frac{\angle ABC}{2} \Rightarrow \angle EAB + \angle ABE = \frac{\angle DAB + \angle ABC}{2}</math> and <math>\angle DAB + \angle ABC = 180^\circ \Rightarrow \angle EAB + \angle ABE = 90^\circ \Rightarrow \angle AEB = 90^\circ.</math> | ||
− | By <math>1</math> and <math>2,</math> we can conclude that <math>HFGE</math> is a rectangle. <math> | + | By <math>1</math> and <math>2,</math> we can conclude that <math>HFGE</math> is a rectangle. |
+ | |||
+ | Now, knowing <math>HFGE</math> is a rectangle, we can continue on. | ||
+ | |||
Let <math>AB = a, BC = b, </math> and <math>\angle ABE = \alpha.</math> Thus, <math>[ABCD] = ab\sin(2\alpha).</math> | Let <math>AB = a, BC = b, </math> and <math>\angle ABE = \alpha.</math> Thus, <math>[ABCD] = ab\sin(2\alpha).</math> | ||
<math>AD \parallel DC \Rightarrow \angle BJC = \angle JCD</math> and <math>\angle JCD = \angle JCB \Rightarrow \angle BJC = \angle JCB \Rightarrow JB = BC =b.</math> | <math>AD \parallel DC \Rightarrow \angle BJC = \angle JCD</math> and <math>\angle JCD = \angle JCB \Rightarrow \angle BJC = \angle JCB \Rightarrow JB = BC =b.</math> |
Revision as of 16:22, 22 October 2020
The bisectors of the internal angles of parallelogram with
determine a quadrilateral with the same area as
. Determine, with proof, the value of
.
Solution 1
We claim the answer is Let
be the new quadrilateral; that is, the quadrilateral determined by the internal bisectors of the angles of
.
Lemma :
is a rectangle.
is a parallelogram.
as
bisects
and
bisects
By the same logic,
is a parallelogram.
2.
and
and
By
and
we can conclude that
is a rectangle.
Now, knowing is a rectangle, we can continue on.
Let and
Thus,
and
By the same logic,
and
Because
we have
Solution by Sp3nc3r