Isosceles trapezoid

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Isosceles trapezoid

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Source: Argentina MO 2009

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#1 h Nov 10, 2009, 12:35 PM • 2 Y

Y by Adventure10, Mango247

Isosceles trapezoid $ABCD$ , with $AB \parallel CD$ , is such that there exists a circle $\Gamma$ tangent to its four sides. Let $T = \Gamma \cap BC$ , and $P = \Gamma \cap AT$ ( $P \neq T$ ).
If $\frac{AP}{AT} = \frac{2}{5}$ , compute $\frac{AB}{CD}$ .

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#2 h Nov 11, 2009, 8:04 PM • 1 Y

Y by Adventure10

My answer is 13/3.Is this correct?

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#3 h Jun 14, 2010, 6:41 PM • 2 Y

Y by Adventure10, Mango247

From the given relation, if $AP=2m$ , then $AT=5m$ , $AB=2BT=2m \sqrt{10}$ , which gives $m(\widehat{ABC})=60^\circ$ , hence $AB=2CD$ .

Best regards,
sunken rock

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#4 h Jun 15, 2010, 5:07 AM • 2 Y

Y by Adventure10, Mango247

Actually , i cannot find any pure geometric method to this problem , i would very appreicate any method related !

Here is mine , trigonometric , obviously .

Let AP = 2k , AT = 5k then BT = sqrt((2k)(5k)) = sqrt(10) k and then AB = 2sqrt(10)k

so we have a triangle ABT with side-ratio , AB : BT : TA = 2sqrt(10) k: sqrt(10) k: 5k = 2sqrt(2) : sqrt(2) : sqrt(5) , by law of cosine , we have cos(ABT) = ( 8+2-5)/8 = 5/8 , then tan^2( ABT/2) = ( 1 - 5/8)/(1+5/8) = 3/13 .

AB : CD = 1/(tan^2(ABT/2)) = 13/3 .

In general , if AP : AT = a:b , here 2:5 , then AB : CD = (9a-b):(b-a)

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#5 h Jul 30, 2010, 12:47 AM • 2 Y

Y by Adventure10, Mango247

uglysolutions wrote:

Isosceles trapezoid $ABCD$ , with $AB \parallel CD$ , is such that there exists a circle $\Gamma$ tangent to its four sides. Let $T = \Gamma \cap BC$ , and $P = \Gamma \cap AT$ ( $P \neq T$ ).
If $\frac{AP}{AT} = \frac{2}{5}$ , compute $\frac{AB}{CD}$ .

Let $M,N, R$ the points where $\Gamma$ touch $DA,AB, CD$ , respectively, and let $I$ the center of $\Gamma$ . Because $ABCD$ is an isosceles trapezoid with bases $AB,CD$ , it follows that $AM=AN=BN=BT$ and $DM=DR=CR=CT$ .

Assume that $AP=2a$ , $AT=5a$ for a fix $a\in \mathbb {R}^{+}$ . We have that $AN^2=AP\cdot AT\Rightarrow AN=a\cdot \sqrt{10}$ . Then $AB=2a\cdot \sqrt{10}$ and $BP=a\cdot \sqrt{10}$ . Let $Q=AP\cap DC$ . Note that $\triangle APB\sim \triangle QPC$ (*), and in consequence $CQ=2\cdot CP=CD$ . Assume that que $CD=b$ (where $b\in \mathbb {R}^+$ ). By the similitude in (*) $PQ=b\cdot \sqrt{5/8}$ . Then $QR^2=QT\cdot QP$ , that is:

$\dfrac{9}{4}b^2=b\cdot \sqrt{\dfrac{5}{8}}\cdot \left(b\cdot \sqrt{\dfrac{5}{8}}+3a\right)$

It follows that $b=\dfrac{3a}{13}\cdot \sqrt{40}$ . Because $AB=2\cdot AN=a\cdot \sqrt{40}$ , we can conclude that: $\dfrac{AB}{CD}=\dfrac{13}{3}$ . $\blacksquare$

I have posted this solution in http://www.fmat.cl forum

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