Geometry Question

[Dancer25]

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!

 

[Denis]

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

 

[Dancer25]

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?

 

[BigGlenntheHeavy]

\(\displaystyle Dancer25 \ don't \ feel \ bad, \ I'm \ stuck \ also.\)

\(\displaystyle A \ little \ more \ elucidation \ Denis?\)

 

[Dancer25]

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

 

[BigGlenntheHeavy]

\(\displaystyle Really, \ how \ about \ showing \ it \ to \ the \ rest \ of \ us.\)

 

[Dancer25]

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

 

[soroban]

Hello, Dancer25!

Your work is correct . . . Nice going!

Here's another approach . . .

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

\(\displaystyle \text{Find }(AC)^\)

\(\displaystyle \text{The correct answer is 90, I believe.}\)

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

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


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


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

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


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


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

 

[Denis]

Go drink a couple of coffees, BigGlen :wink:

 

[BigGlenntheHeavy]

\(\displaystyle Good \ show \ soroban, \ as \ the \ circle \ never \ came \ into \ play \ in \ your \ analysis, \ which \ was \ confusing\)

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

\(\displaystyle Oh, \ thanks \ for \ nothing \ Denis.\)

 

[Deleted member 4993]

\(\displaystyle Good \ show \ soroban, \ as \ the \ circle \ never \ came \ into \ play \ in \ your \ analysis, \ which \ was \ confusing\)

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

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

 

[BigGlenntheHeavy]

\(\displaystyle You \ are \ right \ Subhotosh \ Khan, \ I \ stand \ corrected.\)

\(\displaystyle A \ cyclic \ quadrilateral, \ I'll \ remember \ this \ one.\)

\(\displaystyle Somehow \ I \ always \ thought \ that \ the \ diagonal \ opposite \ angles \ of \ any \ quadrilateral\)

\(\displaystyle were \ supplementary, \ live \ and \ learn.\)

 

[Denis]

Go drink a couple of coffees, BigGlen :wink:

Go have 2 more :idea: