Chord Distance

Let $\angle ADB$ be $x$ . We have $BD=\sqrt{400^2-60^2}=10\sqrt{1564}$ . So, $\sin x=\frac{60}{400}$ and $\cos x=\frac{10\sqrt{1564}}{400}\Rightarrow \sin 2x=2\sin x\cos x=2\cdot \frac{60}{400}\cdot \frac{10\sqrt{1564}}{400}=\frac{3\sqrt{1564}}{400}$ .

$AC=\sin 2x \cdot AD=\frac{3\sqrt{1564}}{400}\cdot 400=3\sqrt{1564}$ . $CD=\sqrt{400^2-(3\sqrt{1564})^2}=382$ . Done.

First, scale down the dimensions by dividing everything by 20. Let the center of the circle be O. using cosine law for triangles $AOB$ and $BOC$ , we get $\cos \angle AOB = \cos \angle BOC = \frac {191}{200}$ . Since $AD$ is a diameter, $\angle COD$ = $180^\circ - 2 \angle AOB$ . Applying cosine law to $COD$ , $\ |CD|^2=200-200cos \angle COD=200+200 \times \cos 2\angle AOB$ $=200+200 \times (2 \frac{191^2}{200^2} -1)= \frac{191^2}{100}$ . Hence, $|CD|=\frac{191}{10}$ . To bring it back to original scale, multiply by 20. Therefore, $|CD|=382$ .

If we note that ABD and ACD are right triangles and use Ptolemy's Theorem on quad. ABCD (being cyclic), we get a system:

60 CD + 60(400) = (AC)(BD)

400^2 = 60^2 + BD^2

400^2 = CD^2 + AC^2

Squaring the first equation and substituting the values of BD^2 and AC^2 from the second and third equations, respectively, we get an equation entirely in CD:

[60(CD+400)]^2 = AC^2 BD^2 = (400^2-CD^2)(400^2-60^2)

60^2(400+CD)(400+CD) = (400+CD)(400-CD)(400^2-60^2).

We can cancel out the common factor 400+CD to get a linear equation in CD:

60^2(400) + 60^2(CD) = 400(400^2-60^2)

CD = [400^3-60^2(400) - 60^2(400)]/400^2

=400-60^2/400-60^2/400

=400-9-9

=382

Let $\angle ADB$ be $x$ . We have $BD=\sqrt{400^2-60^2}=10\sqrt{1564}$ . So, $\sin x=\frac{60}{400}$ and $\cos x=\frac{10\sqrt{1564}}{400}\Rightarrow \sin 2x=2\sin x\cos x=2\cdot \frac{60}{400}\cdot \frac{10\sqrt{1564}}{400}=\frac{3\sqrt{1564}}{400}$ .

$AC=\sin 2x \cdot AD=\frac{3\sqrt{1564}}{400}\cdot 400=3\sqrt{1564}$ . $CD=\sqrt{400^2-(3\sqrt{1564})^2}=382$ . Done.

As in any geometry problem, a good diagram is crucial. In this case, a diagram reveals a cyclic quadrilateral (a quadrilateral inscribed in a circle) ABCD. We are given $AB = 60, BC = 60, AD=400$ and we desire $CD=x$ . First, Ptolemy's theorem states that for a cyclic quadrilateral, $(AB)(CD) + (BC)(AD) = (AC)(BD)$ . If we had an expression for the two diagonals in terms of $x$ , we could solve the equation and get CD.

To find these expressions, recall that a triangle formed with one side as the diameter and a point on the circumference is a right triangle. Thus, triangles $ABD$ and $ACD$ are right, with $B$ and $D$ the right angles respectively. Applying the pythagorean theorem to $ABD$ , we get $(AB)^2 + (BD)^2 =(AD)^2$ . Substituting in known values, we get $60^2 + (BD)^2 = 400^2, or BD = (156400)^.5 = 20(391)^.5$ Similarly, for Triangle ACD, we get an expression involving $x$ : $(AC)^2 +(CD)^2 = (AD)^2 (AC) = (400^2 - x^2).5$ . With all the lengths known or in terms of x, we are ready to simply plug into Ptolemy's Theorem:

$(AB)(CD) + (BC)(AD) = (AC)(BD) (60)(x) + (60)(400) = (20(391)^.5)(400^2-x^2)^.5 .$

To get rid of radicals, divide both sides by 20 and square: $9x^2 + 7200x + 1200^2 = 391(400)^2 - 391x^2$ . Bring all terms to the left side, and divide by 400: $x^2 + 18x - 152800$ . We are now in a position to use the quadratic formula: $x = (-18+(18^2-4(-152800)^.5)/2$ , $x=382$ . Thus, $x = CD= 382$ .

[Note: Since we squared the equation, we should ensure that we didn't introduce 382 as a solution, and have to verify that it indeed works. In the case, the introduced solution was $-400$ . - Calvin]

Quadrilateral ABCD is cyclic and we know the diagonals AB and BD can be found using the Pythagorean Theorem because angles ACD and ABD are right (they are inscribed in the diameter). Thus, we can set up Ptolemy's Theorem as follows: $60 \times 400 + 60 \times |CD| = (\sqrt{400^2-60^2}) \times (\sqrt{400^2-|CD|^2})$

$\sqrt{400^2-60^2}$ factors into $20 \times \sqrt{391}$ from which we can factor out 20 giving us an equation of $1200+3 \times |CD|=\sqrt{391} \times \sqrt{400^2-|CD|^2}$ From here, we can square both sides to get $1200^2+7200 \times |CD| +9 \times |CD|^2=391 \times (400^2-|CD|^2)$ and then move like terms to get a quadratic and divide by 400 giving $|CD|^2+18 \times |CD|+3600-391 \times 400=0$ We can solve this large quadratic by factoring out $4$ from the discriminant $18^2-4 \times (3600-391 \times 400)$ and realizing that $4 \times (18^2-4 \times (3600-391 \times 400))=4 \times 391^2$ . Thus, the answer is $\frac{-18+2 \times 391}{2}=382$

Draw diagonals BD and AC.

Now, from Ptolemy's Theorem, |AC| * |BD| = |BC| * |AD| + |AB| * |CD|.

Substituting, we have |AC| * |BD| = 60 * 400 + 60 * |CD| = 60 * (400 + |CD|).

We have angle ABD = angle ACD = 90 degrees, so |BD|^2 = 400^2 - 60^2 = 156400 and |AC|^2 = 400^2 - |CD|^2, by the Pythagorean Theorem.

Now, 156400 * (400^2 - |CD|^2) = |AC|^2 * |BD|^2 = 60^2 * (400 + |CD|)^2.

So we have, 156400 * (160000 - |CD|^2) = 3600 * (400 + |CD|)^2.

The resulting equation is a quadratic equation with roots -400 and 382. Since |CD| is positive, |CD| = 382.

Ptolemy's Theorem states that in any cyclic quadrilateral, the sum of the products of opposite sides equals the product of the diagonals. Quadrilateral ABCD is cyclic because it is on Now, (AB)(CD)+(BC)(AD)=(BD)(AC) Since AB=BC=60 and AD=400 LHS becomes 60(400+CD)

We can write the RHS in terms of CD using the Pythagorean theorom. Triangles ABD and ACD are inscribed in a semicircle, so they are right angled. BD^2=400^2-60^2 BD=sqrt(156400)

and AC^2=400^2-CD^2 AC=sqrt(160000-CD^2)

Finally, we have the equality, 60(400+CD)=sqrt((156400)(160000-CD^2))

Solving this and taking the positive of the two roots gives the final answer of CD=382

ABCD is a cyclic quadrilateral in a cyclic quadrilateral BC AD + AB CD = AC BD (PTOLEMYS THEOREM) AC and BD can be found using pythagoras theorem as in a semicircle the angle subtended by the diameter to any point is 90 degrees now equation reduces to 60 400 + 60 CD = {[400^2-CD^2]^1/2} {[400^2-60^2]^1/2} solving CD=382

Let $O$ be the center of the circle, and let $P$ be the midpoint of $AC$ . Since $|AB|=|BC|$ , $BP$ is a perpendicular bisector of $AC$ . Since $|OA|=|OC|$ , $OP$ is a perpendicular bisector of $AC$ . Thus, $O, P$ and $B$ lie on a straight line. Since $P$ is a midpoint of $AC$ and $O$ is a midpoint of $AD$ , thus $PO \parallel CD$ and $|PO| = \frac{|CD|}{2}$ . We let $|CA| = 2x$ and $|CD|=2y$ . Then from right triangle $CPO$ , Pythagorean theorem gives $|CP|^2 + |PO|^2 = |CO|^2 \Rightarrow x^2+y^2 = 200^2$ . Also, from right triangle $CPB$ , Pythagorean theorem gives $|CP|^2 + |PB|^2 = |CB|^2 \Rightarrow x^2 + (200-y)^2 = 60^2$ . Taking the difference of these equations, we get $400y - 200^2=200^2 - 60^2$ $\Rightarrow y = \frac {200^2+200^2-60^2}{400}=191$ . Thus $|CD|=2y = 2\times 191 = 382$ .

First, draw a circle and the points A, B, C, D approximately so the problem conditions are met. Then draw \overline{AC}. Draw a radius to point B and then chord \overline{AC}. Label the center as O and the intersection of \overline{AC} and \overline{BO} as point E. Triangles ABE and AEO are right triangles, so a relationship between \overline{BE} and (overline{EO}) can be related through the radius 200. Find \overline{AE} and double it to get AC. Since triangle AED is a right triangle, use the Pythagorean theorem to find \overline{CD}. The solution is 382.

Wise use of $\large Cos(C)=\frac{a^2+b^2-c^2}{2ab}$ is already enough to solve the problem.

The center of the circle is the point O. The radius is |AO|=|OD|=|BO|=|CO|=400/2=200. There are 3 isosceles triangles in the circle: ABO, BCO, CDO. We can use the Law of the Cosines on them. On ABO: AB^2=|AO|^2+|BO|^2-2 |AO| |BO| cosAÔB. Substituting the values given in the exercise, we can find cosAÔB, and it's equal to 0.955; arccos0.955 is equal to approximately 17.25 grades. It is the same value of BÔC, because |AB|=|BC|, and so triangle ABO is equal to triangle BCO. As the arc |AD| measures 180 grades, and the minor arc |AC| measures AÔB+BÔC=17.25+17.25=34.5 grades, approximately, the minor arc CÔD=180-34.5=145.5 grades, approximately. We can use the Law of Cosines in triangle CDO: |CD|^2=|CO|^2+|OD|^2-2 |CO| |OD| cos145,5. Hence the lenght |CD| is approximately equal to 382.

First let $\angle AOB$ be $x$ . Then we use the cosine law on $\triangle AOB$ : $60^2=200^2+200^2-2*200*200*cos(x)$ . Solving we get $cos(x)=\frac {191}{200}$ . Notice $\angle ADC$ is half of $\angle AOC$ due to $Central Angle Theorem$ thus $\angle AOB=\angle ADC$ . Now we set $CD=y$ and apply the cosine law on $\triangle OCD$ , getting $200^2=200^2+y^2-2*200*y*cos(x)$ . Substituting $\frac{191}{200}$ into $cos(x)$ and solving we get $a=0$ and $a=382$ . Since $a$ cannot equal to $0$ , the answer is $\boxed{382}$

While I can't really draw it, imagine the problem as described with radii from the center of the circle O to B and C. From here, you can see how angle AOB is congruent to angle BOC and that their combined angle AOC is supplementary to angle COD. using the law of cosines, we find that $-\frac{(60^2 - 200^2 - 200^2)}{(2 * 200 * 200)} = \cos \theta$ where $\theta$ is half of m $\angle AOC$ .

We can now find m $\angle AOC$ by simple plug and chug, but we'll wait on that. since we know AOC is supplementary to COD, we can find COD and plug it into the law of cosines once again to find that CD is... presses buttons on calculator ...382!