IOQM 2022 P5

Art of Problem Solving

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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

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IOQM 2022 P5

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1184 posts

L567

#1 h Mar 8, 2022, 8:42 AM • 1 Y

Y by Master_of_Aops

In parallelogram $ABCD$ , the longer side is twice the shorter side. Let $XYZW$ be the quadrilateral formed by the internal bisectors of the angles of $ABCD$ . If the area of $XYZW$ is $10$ , find the area of $ABCD$

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Numbertheorydog

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Numbertheorydog

#2 h Mar 8, 2022, 11:02 AM • 2 Y

Y by icecream1234, DeathHammer7

The key is that the angle bisectors meet on the midpoint of the opposite sides.
and also form a rectangle.
from here it is easy

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mathlearner2357

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mathlearner2357

#3 h Mar 8, 2022, 12:22 PM • 2 Y

Y by Numbertheorydog, icecream1234

i just assumed that the parallelogram is a rectangle and from there it is easy

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I_love_chom-choms

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I_love_chom-choms

#4 h Mar 8, 2022, 1:05 PM

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@numbertheorydog @mathlearner2357
even if you dont, its still a simple question, you just have to apply BPT which gives you the area

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656 posts

#5 h Jun 14, 2022, 6:24 PM

Y by

This seems similar to USAMTS Year 32 Round 1 Problem 3.

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#6 h Oct 20, 2022, 6:21 PM • 1 Y

Y by DeathHammer7

I will just post a solution because I am mastermind who blundered it in exam even after seeing the bash on exam.
Let $\overline{AB}$ be the longer side of the parallelogram. Let $\angle DAB = 2\theta$ . One can angle chase to find that $XYZW$ is a rectangle. Let $B$ bisector meet $C$ bisector at $X$ and $Y$ be the intersection of $A$ bisector and $B$ bisector.

From $\triangle XBC$ , $\overline{XB} = x\sin(\theta)$ . From $\triangle AYB$ , $\overline{XY} = 2x\sin(\theta)$ . With $\overline{XB}$ and $\overline{XY}$ . one can deduce $X$ is the midpoint of $\overline{YB}$ . Now one can compute side length of the rectangle as $x\sin(\theta)$ . Similarly, one can find the other one to be $x\cos(\theta)$ . By the problem condition
$x\sin(\theta) \cdot x\cos(\theta) = 10 \iff \sin(2\theta) = \frac{20}{x^2}$ Now $2 [\triangle BCD]$ is the area of parallelogram. One can just use the $\frac{1}{2}ab\sin(C)$ formula now to get
$\text{Desired Area} = 2 \times \frac{1}{2}x\cdot 2x \cdot \sin(2\theta) = \fbox{40}$ and thus we are done. $\blacksquare$

This post has been edited 2 times. Last edited by SatisfiedMagma, Oct 20, 2022, 6:22 PM

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Tutan

#7 h Sep 1, 2023, 4:40 PM

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Draw ABCD parallelogram
We know that internal angle bisectors of the angles in a parallelogram meets at the mid point of the opposite side and it forms a rectangle.
now if we draw the XYZW rectangle and denote the two intersections Y and W then the line passes through X,Z is parallel to AD and BC
Let the line cut AB and DC at M and N
Now (XYZ)=(XWZ)=5
(XYZ)=1/2(YXZD)=1/2(YZXA)
(AXY)=(YZD)=5
(AMND)=(AWD)=5X4=20
(ABCD)=2(AMND)=40(ans.)
Note:- (F)denotes the area of geometrical figure F.

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359 posts

#8 h Sep 1, 2023, 6:16 PM

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Numbertheorydog wrote:

The key is that the angle bisectors meet on the midpoint of the opposite sides.
and also form a rectangle.
from here it is easy

Indeed, by splitting the parallelogram into two congruent rhombuses, we can see that the angle bisectors split these rhombuses into 4 congruent triangles each. The rest is trivial.

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