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by **Dancer25** » Mon Jul 19, 2010 3:16 am

Hey everyone, I'm having a bit of blank on a seemingly easy geometry question. Could someone please give some mathematical insight? "Quadrilateral ABCD is inscribed in a circle. AB=5, BC=5, CD=5root2, AD=10root2. Find (AC)^2." The correct answer is 90, I believe. Thanks!

by **Denis** » Mon Jul 19, 2010 3:44 am

AC / BD = (AB·AD + BC·CD) / (AB·BC + AD·CD).

by **Dancer25** » Tue Jul 20, 2010 5:25 am

Hmm... I'm still having a bit of trouble. I got that AC/BD is 3root2/5, and by similarity I got that if we let the point at which the diagonals intersect be P, then 2(CP)= AP. It's probably something obvious, but what is the final step?

by **BigGlenntheHeavy** » Tue Jul 20, 2010 5:50 am

by **Dancer25** » Tue Jul 20, 2010 5:51 am

Ah, nevermind. I used Ptolemy's and found the answer.

by **BigGlenntheHeavy** » Tue Jul 20, 2010 5:54 am

by **Dancer25** » Tue Jul 20, 2010 6:09 am

Haha okay, here's the mess of what I did.

I then noticed that triangles ABP and PCD are similar (AA similarity) .

Let AP= x, PC= y, BP = w, PD =z. Using ratios, we get that x= (root2{z})\2, and y=root2(w).

Using the extension of Ptolemy's Denis suggested- x+y/w+z = 3root2/5.

If we substitute the values obtained above into this equation- y= 2x.

Now I used Ptolemy's [(AC)(BD) = (AB)(CD) + (BC)(AD)] in terms of y...

(3y)({[5root2]y}\ 2)= 75root 2... y=root10.

Thus, x=2root10, x+y = 3root10, and (AC)^2 = 90

by **soroban** » Tue Jul 20, 2010 1:43 pm

Hello, Dancer25!

Your work is correct . . . Nice going!

Here's another approach . . .

$\text{Cyclic quadrilateral }ABCD\text{ with: }\:AB=5,\;BC=5,\;CD=5\sqrt{2},\;AD=10\sqrt{2}$

$\text{Find }(AC)^$

$\text{The correct answer is 90, I believe.}$

$\text{Using the Law of Cosines on }\Delta ABC:\;\;AC^2 \:=\:5^2 + 5^2 - 2(5)(5)\cos B\;\;[1]$

$\text{Using the Law of Cosines on }\Delta ADC:\;\;AC^2 \:=\:(5\sqrt{2})^2 + (10\sqrt{2})^2 - 2(5\sqrt{2})(10\sqrt{2})\cos D\;\;[2]$

$\text{Equate [1] and [2]: }\;50 - 50\cos B \;=\;250 - 200\cos D \quad\Rightarrow\quad 4\cos D - \cos B \:=\:4\;\;[3]$

$\text{Note that }\angle B\text{ and }\angle D\text{ are supplementary: }\;B + D \:=\:180^o \quad\Rightarrow\quad D \:=\:180^o - B$

. . $\text{Hence: }\:\cos D \:=\:\cos(180^o-B) \:=\:-\cos B$

$\text{Then [3] becomes: }\;-4\cos B - \cos B \:=\:4 \quad\Rightarrow\quad -5\cos B \:=\:4 \quad\Rightarrow\quad \cos B \:=\:-\tfrac{4}{5}$

$\text{Substitute into [1]: }\;AC^2 \:=\:50 - 50\left(-\tfrac{4}{5}\right) \;=\;90$

by **Denis** » Tue Jul 20, 2010 2:37 pm

Go drink a couple of coffees, BigGlen :wink:

by **BigGlenntheHeavy** » Tue Jul 20, 2010 6:09 pm

by **Subhotosh Khan** » Tue Jul 20, 2010 10:27 pm

BigGlenntheHeavy wrote: $Good \ show \ soroban, \ as \ the \ circle \ never \ came \ into \ play \ in \ your \ analysis, \ which \ was \ confusing$

$me. \ In \ fact, \ I \ did \ it \ your \ way, \ but \ thought \ the \ circle \ had \ something \ to \ do \ with \ it.$

The circle does come into play - that is why angles ABC and ADC are supplementary.

by **BigGlenntheHeavy** » Wed Jul 21, 2010 1:40 am

by **Denis** » Wed Jul 21, 2010 4:26 am

Denis wrote:Go drink a couple of coffees, BigGlen

Go have 2 more :idea:

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