Area, Perimeter and Quadrilateral

Geometry Level 5

This problem is from the OMO.

Let $ABCD$ be a quadrilateral with $AD = 20$ and $BC = 13$ . The area of $ABC$ is $338$ and the area of $DBC$ is $212$ . Compute the smallest possible perimeter of $ABCD$ .

The answer is 118.

2 solutions

Zi Song Yeoh
Sep 17, 2014

Let $X$ and $Y$ be the feet of the altitudes from $A$ and $D$ to $BC$ . By the area condition, $AX = \frac{676}{13}$ and $DY = \frac{424}{13}$ . Let $Z$ be a point such that $AXYZ$ form a rectangle. Thus, $XY = AZ = \sqrt{AD^2 - (ZY - DY)^2} = \sqrt{AD^2 - (AX - DY)^2} = \sqrt{400 - (\frac{252}{13})^2} = \frac{64}{13}$ .

Let $E$ be the reflection of $D$ over $Y$ and $F$ be the point such that $BCEF$ is a parallelogram. Then it’s easy to compute that

$AF^2 = (XY - BC)^2 + (AX + DY)^2 = (\frac{105}{13})^2 + (\frac{1100}{13})^2 = 7225 = 85^2$ .

Thus, $AB + CD = AB + CE = AB + BF \ge AF = 85$ , with equality if and only if $A, B, F$ are collinear. Thus, the minimum possible perimeter of $ABCD$ is $85 + 20 + 13 = \boxed{118}$ .

Hey Zi Song Yeoh Check This:

$AB=y\quad \\ BC=13\\ CD=x\\ AD=20$ .

Now By using given Area conditions we obtain Relation for 'x' and 'y'

$x=\frac { 424 }{ 13\sin { C } } \\ y=\frac { 52 }{ \sin { B } }$ .

Now Perimeter of Quadrilateral ABCD is

$P=\quad 13+20+x+y\\ P=\quad 33+\frac { 424 }{ 13\sin { C } } +\frac { 52 }{ \sin { B } } \\$ .

For Minimum Perimeter Denominator containing angles should Be maximum. Since

${ \sin { B } }_{ max }\quad =\quad { \sin { C } }_{ max }\quad =\quad 1\\$ .

which ocures when $B=C={ 90 }^{ 0 }$ .

i.e Quadrilateral becomes Trapazium.

${ P }_{ min }=\quad 117.6$ .

From my side this is minimum value . Please Correct me if I'am wrong.

Deepanshu Gupta - 6 years, 8 months ago

Can you define the quadrilateral exactly? Just because you gave some side lengths and some angles, doesn't necessarily mean that you have defined a quadrilateral.

As an explicit example of what I mean, there is no quadrilaterial where $AB = 1, BC = 2, CD = 3, DA = 4, \angle ABC = 90^\circ, \angle BCD = 90 ^ \circ$ .

The reason is that this system is overdefined, and there is a contradiction somewhere.

Calvin Lin Staff - 6 years, 8 months ago

Now I get it...!! Since to make an Right angle Trapezium in my solution Condition is That $x-y+13>20$ . (According to triangle inequality in a triangle which is formed from by droping the perpendicular from A to CD ).

Which is Violated if I assumed That angle B =angle C =90 degree.

Since x=424/13 , y= 52 , which Don't Satisfy The inequality.

Hence Quadrilateral is not formed..... Am I right ??

Deepanshu Gupta - 6 years, 8 months ago

Rahul Dandwate
Oct 8, 2014

Let $X$ and $Y$ be the feet of perpendicular from $A$ and $D$ to $BC$ . By using the area of triangles given we get $AX=52$ and $DY=\frac { 424 }{ 13 }$ . Let $Z$ be the foot of perpendicular from $D$ to $AB$ . Thus,

$DZ=XY=\sqrt { { AD }^{ 2 }-{ (AX-DY) }^{ 2 } } =\sqrt { 400-{ (\frac { 252 }{ 13 } ) }^{ 2 } } =\frac { 64 }{ 13 }$

Let $\angle BAX=\beta$ and $\angle CDY=\alpha$ . And $BX={ l }_{ 1 }$ and $CY={ l }_{ 2 }$ . Therefore, we get

$13=\frac { 64 }{ 13 } +{ l }_{ 1 }+{ l }_{ 2 }$ $\Longrightarrow \frac { 105 }{ 13 } ={ l }_{ 1 }+{ l }_{ 2 }$

As ${ l }_{ 1 }=52\tan { \beta }$ and ${ l }_{ 2 }=\frac { 424 }{ 13 } \tan { \alpha }$ .Thus,

$52\tan { \beta } +\frac { 424 }{ 13 } \tan { \alpha } =\frac { 105 }{ 13 }$

We have $AB=52\sec { \beta }$ and $CD=\frac { 424 }{ 13 } \sec { \alpha }$ . Now we want to find the minimum value of $AB+CD$ . Using Lagrange Multiplier,

$\Lambda (\alpha ,\beta ,\lambda )=52\sec { \beta } +\frac { 424 }{ 13 } \sec { \alpha } +\lambda (52\tan { \beta } +\frac { 424 }{ 13 } \tan { \alpha } -\frac { 105 }{ 13 } )$

We get $-\lambda =\sin { \alpha } =\sin { \beta }$ . Therefore, $\alpha =\beta$ (as $\alpha <90$ and $\beta <90$ ).

Now solving for $\alpha$ and then using it, we find $AB+CD=85$

Hence, the minimum perimeter of the quadrilateral is $20+13+85=\boxed{118}$ .

Area, Perimeter and Quadrilateral

The answer is 118.

2 solutions

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