1983 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If $x\ne0,$ $\frac{x}{2}=y^2$ and $\frac{x}{4}=4y,$ then $x$ equals

$8$

$16$

$32$

$64$

$128$

Concepts:system of equations substitution zero product property

Difficulty rating: 1380

Small Hint:

Use $\frac{x}{4}=4y$ to express $x$ in terms of $y$

Big Hint:

After substitution, discard the solution that makes $x=0$

Solution:

The second equation gives $x=16y.$ Substituting into the first gives $8y=y^2,$ so $y(y-8)=0.$ The condition $x\ne0$ rules out $y=0,$ leaving $y=8$ and $x=128.$

Therefore, the correct answer is E.

2.

Point $P$ is outside circle $C$ on the plane. At most how many points on $C$ are $3$ cm from $P?$

$1$

$2$

$3$

$4$

$8$

Concepts:circle counting intersections

Difficulty rating: 1130

Small Hint:

Points at a fixed distance from $P$ form another circle

Big Hint:

Count the common points of the two circles; the maximum occurs when they cross

Solution:

The points $3$ cm from $P$ form a circle centered at $P.$ Two distinct circles intersect in at most two points, and two intersections are possible.

Therefore, the correct answer is B.

3.

Three primes, $p,$ $q$ and $r,$ satisfy $p+q=r$ and $1\lt p\lt q.$ Then $p$ equals

$2$

$3$

$7$

$13$

$17$

Concepts:prime parity

Difficulty rating: 1260

Small Hint:

Every prime except one is odd

Big Hint:

If both addends were odd, their sum would be an even number greater than $2$

Solution:

If both $p$ and $q$ were odd, then $r=p+q$ would be even and greater than $2,$ so it would not be prime. Thus one addend is the only even prime, $2.$ Since $p\lt q,$ that addend is $p.$

Therefore, the correct answer is A.

4.

In the adjoining plane figure, sides $AF$ and $CD$ are parallel, as are sides $AB$ and $FE,$ and sides $BC$ and $ED.$ Each side has length $1.$ Also, $\angle FAB=\angle BCD=60^\circ.$ The area of the figure is

$\frac{\sqrt3}{2}$

$1$

$\frac32$

$\sqrt3$

$2$

Concepts:area area decomposition equilateral triangle

Difficulty rating: 1660

Small Hint:

Place $AF$ and $CD$ vertically and use the $60^\circ$ angles to assign coordinates

Big Hint:

The six vertices can be written using horizontal changes of $\frac{\sqrt3}{2}$

Solution:

The parallel directions and $60^\circ$ angles place the vertices, up to rigid motion, at $\begin{aligned} A&=(0,0),\\ B&=(\tfrac{\sqrt3}{2},-\tfrac12),\\ C&=(\sqrt3,0),\\ D&=(\sqrt3,-1),\\ E&=(\tfrac{\sqrt3}{2},-\tfrac32),\\ F&=(0,-1). \end{aligned}$ Applying the shoelace formula to these six vertices gives area $\sqrt3.$

Therefore, the correct answer is D.

5.

Triangle $ABC$ has a right angle at $C.$ If $\sin A=\frac23,$ then $\tan B$ is

$\frac35$

$\frac{\sqrt5}{3}$

$\frac{2}{\sqrt5}$

$\frac{\sqrt5}{2}$

$\frac53$

Concepts:trigonometry right triangle angle sum

Difficulty rating: 1330

Small Hint:

Angles $A$ and $B$ are complementary

Big Hint:

Use $\cos A=\sqrt{1-\sin^2A}$ and $\tan B=\cot A$

Solution:

Since $\sin A=\frac{2}{3},$ $\cos A=\sqrt{1-\frac{4}{9}}=\frac{\sqrt5}{3}.$ Because $B=90^\circ-A,$ we have $\tan B=\cot A$ and $\frac{\cos A}{\sin A}=\frac{\sqrt5}{2}.$

Therefore, the correct answer is D.

6.

When $x^5,$ $x+\frac1x$ and $1+\frac2x+\frac3{x^3}$ are multiplied, the product is a polynomial of degree

$2$

$3$

$6$

$7$

$8$

Concepts:polynomial exponent

Difficulty rating: 1450

Small Hint:

Track the largest exponent produced by choosing one term from each factor

Big Hint:

The leading term comes from $x^5\cdot x\cdot1$

Solution:

The greatest exponent comes from $x^5\cdot x\cdot1=x^6.$ Every other term has smaller exponent, and the smallest possible exponent is $5-1-3=1,$ so the product is indeed a polynomial. Its degree is $6.$

Therefore, the correct answer is C.

7.

Alice sells an item at $\$10$ less than the list price and receives $10\%$ of her selling price as her commission. Bob sells the same item at $\$20$ less than the list price and receives $20\%$ of his selling price as his commission. If they both get the same commission, then the list price is

$\$20$

$\$30$

$\$50$

$\$70$

$\$100$

Concepts:percentage money linear equation

Difficulty rating: 1160

Small Hint:

Let $L$ be the list price and write each commission in terms of $L$

Big Hint:

Set $0.1(L-10)$ equal to $0.2(L-20)$

Solution:

If the list price is $L,$ equality of commissions gives $0.1(L-10)=0.2(L-20).$ Multiplying by $10$ and solving gives $L-10=2L-40,$ so $L=30.$

Therefore, the correct answer is B.

8.

Let $f(x)=\frac{x+1}{x-1}.$ Then for $x^2\ne1,$ $f(-x)$ is

$\frac1{f(x)}$

$-f(x)$

$\frac1{f(-x)}$

$-f(-x)$

$f(x)$

Concepts:function rational equation algebraic manipulation

Difficulty rating: 1360

Small Hint:

Substitute $-x$ directly into the formula

Big Hint:

Multiply numerator and denominator of $f(-x)$ by $-1$

Solution:

We have $\begin{aligned} f(-x)&=\frac{-x+1}{-x-1}\\ &=\frac{x-1}{x+1}=\frac1{f(x)}. \end{aligned}$ The restriction $x^2\ne1$ makes all displayed quantities defined.

Therefore, the correct answer is A.

9.

In a certain population the ratio of the number of women to the number of men is $11$ to $10.$ If the average (arithmetic mean) age of the women is $34$ and the average age of the men is $32,$ then the average age of the population is

$32\frac9{10}$

$32\frac{20}{21}$

$33$

$33\frac1{21}$

$33\frac1{10}$

Concepts:weighted mean ratio and proportion

Difficulty rating: 1240

Small Hint:

Use groups of $11$ women and $10$ men

Big Hint:

Divide $11\cdot34+10\cdot32$ by the total number of people

Solution:

Using $11$ women and $10$ men, the total of all ages is $11\cdot34+10\cdot32=694.$ Dividing by $21$ people gives $\frac{694}{21}=33\frac1{21}.$

Therefore, the correct answer is D.

10.

Segment $AB$ is both a diameter of a circle of radius $1$ and a side of an equilateral triangle $ABC.$ The circle also intersects $AC$ and $BC$ at points $D$ and $E,$ respectively. The length of $AE$ is

$\frac32$

$\frac53$

$\frac{\sqrt3}{2}$

$\sqrt3$

$\frac{2+\sqrt3}{2}$

Concepts:circle inscribed angle equilateral triangle special right triangle

Difficulty rating: 1660

Small Hint:

Because $AB$ is a diameter, $\angle AEB$ is a right angle

Big Hint:

Triangle $ABE$ is a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle with hypotenuse $AB=2$

Solution:

Since $E$ is on the circle with diameter $AB,$ $\angle AEB=90^\circ.$ Also $\angle ABE=60^\circ$ because $ABC$ is equilateral. Thus $ABE$ is a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle with hypotenuse $AB=2,$ so $AE=\sqrt3.$

Therefore, the correct answer is D.

11.

Simplify $\sin(x-y)\cos y$ ${}+\cos(x-y)\sin y.$

$1$

$\sin x$

$\cos x$

$\sin x\cos2y$

$\cos x\cos2y$

Concepts:trigonometric identity algebraic manipulation

Difficulty rating: 1240

Small Hint:

Recognize the sine addition identity

Big Hint:

Apply the sine addition formula with $u=x-y$ and $v=y$

Solution:

Taking $u=x-y$ and $v=y,$ the sine addition identity gives $\begin{aligned} &\sin(x-y)\cos y\\ &\quad+\cos(x-y)\sin y\\ &=\sin((x-y)+y)=\sin x. \end{aligned}$

Therefore, the correct answer is B.

12.

If $\log_7(\log_3(\log_2x))=0,$ then $x^{-\frac{1}{2}}$ equals

$\frac13$

$\frac1{2\sqrt3}$

$\frac1{3\sqrt3}$

$\frac1{\sqrt{42}}$

none of these

Concepts:logarithm exponent

Difficulty rating: 1710

Small Hint:

Undo the logarithms one at a time, starting from the outside

Big Hint:

$\log_7u=0$ implies $u=1,$ then continue inward

Solution:

The equation gives $\log_3(\log_2x)=1,$ so $\log_2x=3$ and $x=8.$ Hence $x^{-\frac{1}{2}}=\frac{1}{\sqrt8}=\frac{1}{2\sqrt2},$ which is not among choices A–D.

Therefore, the correct answer is E.

13.

If $xy=a,$ $xz=b,$ and $yz=c,$ and none of these quantities is $0,$ then $x^2+y^2+z^2$ equals

$\frac{ab+ac+bc}{abc}$

$\frac{a^2+b^2+c^2}{abc}$

$\frac{(a+b+c)^2}{abc}$

$\frac{(ab+ac+bc)^2}{abc}$

$\frac{(ab)^2+(ac)^2+(bc)^2}{abc}$

Concepts:system of equations algebraic manipulation

Difficulty rating: 1730

Small Hint:

Multiply two of $xy=a,$ $xz=b,$ and $yz=c,$ then divide by the third

Big Hint:

Use $x^2=\frac{ab}{c},$ $y^2=\frac{ac}{b},$ and $z^2=\frac{bc}{a}$

Solution:

From the three products, $\begin{aligned} x^2&=\frac{ab}{c},\\ y^2&=\frac{ac}{b},\\ z^2&=\frac{bc}{a}. \end{aligned}$ Putting their sum over denominator $abc$ gives $\begin{aligned} x^2+y^2+z^2 &=\frac{(ab)^2+(ac)^2}{abc}\\ &\quad+\frac{(bc)^2}{abc}. \end{aligned}$

Therefore, the correct answer is E.

14.

The units digit of $3^{1001}7^{1002}13^{1003}$ is

$1$

$3$

$5$

$7$

$9$

Concepts:units digit modular exponentiation

Difficulty rating: 1430

Small Hint:

Reduce $13$ to $3$ when considering units digits

Big Hint:

Combine the two powers of $3$ and use the length- $4$ units-digit cycles

Solution:

Modulo $10,$ $3^{1001}13^{1003}\equiv3^{2004}\equiv1$ because powers of $3$ repeat every $4.$ Also $7^{1002}\equiv7^2\equiv9\pmod {10}.$ Thus the product has units digit $9.$

Therefore, the correct answer is E.

15.

Three balls marked $1,$ $2$ and $3$ are placed in an urn. One ball is drawn, its number is recorded, and then the ball is returned to the urn. This process is repeated and then repeated once more, and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is $6,$ what is the probability that the ball numbered $2$ was drawn all three times?

$\frac1{27}$

$\frac18$

$\frac17$

$\frac16$

$\frac13$

Concepts:conditional probability systematic listing sampling with replacement

Difficulty rating: 1690

Small Hint:

List the ordered triples from $\{1,2,3\}$ whose sum is $6$

Big Hint:

The possibilities are the permutations of $(1,2,3)$ together with $(2,2,2)$

Solution:

The ordered triples with sum $6$ are the six permutations of $(1,2,3)$ and the triple $(2,2,2).$ They are equally likely, and exactly one of the seven has three $2$ ’s. The conditional probability is therefore $\frac{1}{7}.$

Therefore, the correct answer is C.

16.

Let $\begin{aligned} x&=0.123456789101112\\ &\phantom{={}}\ldots998999, \end{aligned}$ where the digits are obtained by writing the integers $1$ through $999$ in order. The $1983$ rd digit to the right of the decimal point is

$2$

$3$

$5$

$7$

$8$

Concepts:digits place value counting integers in a range

Difficulty rating: 1830

Small Hint:

First count the digits contributed by the one- and two-digit integers

Big Hint:

After $189$ digits, locate the remaining position within the three-digit integers

Solution:

The one-digit integers contribute $9$ digits and the two-digit integers contribute $90\cdot2=180,$ for $189$ total. Thus the desired digit is the $1983-189=1794$ th digit among the three-digit integers. Since $1794=598\cdot3,$ it is the last digit of the $598$ th three-digit integer, namely $100+597=697.$ That digit is $7.$

Therefore, the correct answer is D.

17.

The diagram to the right shows several numbers in the complex plane. The circle is the unit circle centered at the origin. One of these numbers is the reciprocal of $F.$ Which one?

$A$

$B$

$C$

$D$

$E$

Concepts:complex number coordinate geometry transformation

Difficulty rating: 1900

Small Hint:

For $z=re^{i\theta},$ its reciprocal has modulus $\frac{1}{r}$

Big Hint:

The reciprocal reflects the argument across the real axis and moves inside the unit circle

Solution:

If $F=re^{i\theta},$ then $\frac{1}{F}=r^{-1}e^{-i\theta}.$ Because $F$ is outside the unit circle in quadrant I, its reciprocal is inside the unit circle in quadrant IV, with the reflected argument. Only point $C$ has those properties.

Therefore, the correct answer is C.

18.

Let $f$ be a polynomial function such that, for all real $x,$ $f(x^2+1)=x^4+5x^2+3.$ For all real $x,$ $f(x^2-1)$ is

$x^4+5x^2+1$

$x^4+x^2-3$

$x^4-5x^2+1$

$x^4+x^2+3$

none of these

Concepts:polynomial substitution algebraic manipulation

Difficulty rating: 1950

Small Hint:

Rewrite the right side as a polynomial in $x^2+1$

Big Hint:

Setting $t=x^2+1$ gives $f(t)=t^2+3t-1$

Solution:

With $t=x^2+1,$ $\begin{aligned} x^4+5x^2+3 &=(t-1)^2\\ &\quad+5(t-1)+3\\ &=t^2+3t-1. \end{aligned}$ Hence $f(t)=t^2+3t-1.$ Substituting $t=x^2-1$ gives $\begin{aligned} f(x^2-1) &=(x^2-1)^2\\ &\quad+3(x^2-1)-1\\ &=x^4+x^2-3. \end{aligned}$

Therefore, the correct answer is B.

19.

Point $D$ is on side $CB$ of triangle $ABC.$ If $\angle CAD=\angle DAB=60^\circ,$ $AC=3$ and $AB=6,$ then the length of $AD$ is

$2$

$2.5$

$3$

$3.5$

$4$

Concepts:angle bisector theorem law of cosines algebraic manipulation

Difficulty rating: 2150

Small Hint:

The angle bisector theorem gives $\frac{DB}{CD}=\frac{AB}{AC}$

Big Hint:

Set $CD=t,$ $DB=2t,$ and $AD=d;$ then apply the law of cosines to the two triangles

Solution:

By the angle bisector theorem, write $CD=t$ and $DB=2t.$ Let $AD=d.$ The law of cosines in triangles $ACD$ and $ABD,$ whose angles at $A$ are both $60^\circ,$ gives $\begin{aligned} t^2&=9+d^2-3d,\\ 4t^2&=36+d^2-6d. \end{aligned}$ Subtracting four times the first equation from the second yields $0=-3d^2+6d.$ Since $d\gt0,$ $d=2.$

Therefore, the correct answer is A.

20.

If $\tan\alpha$ and $\tan\beta$ are the roots of $x^2-px+q=0,$ and $\cot\alpha$ and $\cot\beta$ are the roots of $x^2-rx+s=0,$ then $rs$ is necessarily

$pq$

$\frac1{pq}$

$\frac{p}{q^2}$

$\frac{q}{p^2}$

$\frac pq$

Concepts:Vieta’s Formulas rational equation

Difficulty rating: 1850

Small Hint:

The roots of the second quadratic are reciprocals of the roots of the first

Big Hint:

Use Vieta’s formulas to express their sum and product in terms of $p,q$

Solution:

Let the first roots be $u=\tan\alpha$ and $v=\tan\beta.$ Then $u+v=p$ and $uv=q.$ The second roots are $\frac{1}{u}$ and $\frac{1}{v},$ so $\begin{aligned} r&=\frac1u+\frac1v=\frac pq,\\ s&=\frac1{uv}=\frac1q. \end{aligned}$ Therefore $rs=\frac{p}{q^2}.$

Therefore, the correct answer is C.

21.

Find the smallest positive number from the numbers below

$10-3\sqrt{11}$

$3\sqrt{11}-10$

$18-5\sqrt{13}$

$51-10\sqrt{26}$

$10\sqrt{26}-51$

Concepts:radical inequality rationalizing denominator

Difficulty rating: 1880

Small Hint:

Compare the squares to determine which listed differences are positive

Big Hint:

Rationalize each positive difference using $a-b=\frac{a^2-b^2}{a+b}$

Solution:

Since $100\gt99,$ choice A is positive; since $324\lt325,$ choice C is negative; and since $2601\gt2600,$ choice D is positive while E is negative. The two positive values satisfy $\begin{aligned} 10-3\sqrt{11} &=\frac1{10+3\sqrt{11}},\\ 51-10\sqrt{26} &=\frac1{51+10\sqrt{26}}. \end{aligned}$ The second denominator is larger, so choice D is the smaller positive number.

Therefore, the correct answer is D.

22.

Consider the two functions $\begin{aligned} f(x)&=x^2+2bx+1,\\ g(x)&=2a(x+b), \end{aligned}$ where the variable $x$ and the constants $a$ and $b$ are real numbers. Each such pair of constants $a$ and $b$ may be considered as a point $(a,b)$ in an $ab$ -plane. Let $S$ be the set of such points $(a,b)$ for which the graphs of $y=f(x)$ and $y=g(x)$ do not intersect (in the $xy$ -plane). The area of $S$ is

$1$

$\pi$

$4$

$4\pi$

infinite

Concepts:quadratic circle area inequality

Difficulty rating: 2230

Small Hint:

Set $f(x)=g(x)$ and require the resulting quadratic to have no real roots

Big Hint:

Simplify the negative-discriminant condition in terms of $a$ and $b$

Solution:

Intersections correspond to roots of $x^2+2(b-a)x+(1-2ab)=0.$ There are no real roots exactly when its discriminant is negative: $\begin{aligned} &4(b-a)^2-4(1-2ab)\\ &=4(a^2+b^2-1)\\ &\lt0. \end{aligned}$ Thus $S$ is the interior of the unit circle in the $ab$ -plane, with area $\pi.$

Therefore, the correct answer is B.

23.

In the adjoining figure the five circles are tangent to one another consecutively and to the lines $L_1$ and $L_2.$ If the radius of the largest circle is $18$ and that of the smallest one is $8,$ then the radius of the middle circle is

$12$

$12.5$

$13$

$13.5$

$14$

Concepts:tangent circles similarity geometric sequence

Difficulty rating: 2310

Small Hint:

Every pair of consecutive tangent circles has the same shape after scaling

Big Hint:

The radii form a geometric sequence, so the middle radius squared is the product of the extremes

Solution:

The centers lie on the angle bisector of $L_1$ and $L_2.$ Similarity of the configuration for any two consecutive circles shows that consecutive radii have a constant ratio. Thus the five radii form a geometric sequence. If the middle radius is $m,$ symmetry of a five-term geometric sequence gives $m^2=8\cdot18,$ so $m=12.$

Therefore, the correct answer is A.

24.

How many non-congruent right triangles are there such that the perimeter in cm and area in $\text{cm}^2$ are numerically equal?

none

$1$

$2$

$4$

infinitely many

Concepts:right triangle power scaling of length, area, and volume similarity

Difficulty rating: 2230

Small Hint:

Start with any shape of right triangle and scale all its side lengths by $k$

Big Hint:

Under scaling, perimeter is multiplied by $k$ while area is multiplied by $k^2$

Solution:

Start with any right triangle having perimeter $P$ and area $K.$ Scaling every side by $k=\frac{P}{K}$ produces perimeter $kP$ and area $k^2K,$ which are equal. Infinitely many nonsimilar right-triangle shapes exist, and the resulting triangles are therefore noncongruent.

Therefore, the correct answer is E.

25.

If $60^a=3$ and $60^b=5,$ then $12^{\frac{1-a-b}{2(1-b)}}$ is

$\sqrt3$

$2$

$\sqrt5$

$3$

$\sqrt{12}$

Concepts:logarithm exponent algebraic manipulation

Difficulty rating: 2310

Small Hint:

Express $1-a-b$ and $1-b$ as base- $60$ logarithms

Big Hint:

Use $1-a-b=\log_{60}4$ and $1-b=\log_{60}12$

Solution:

Since $a=\log_{60}3$ and $b=\log_{60}5,$ $\begin{aligned} 1-a-b &=\log_{60}\frac{60}{15}=\log_{60}4,\\ 1-b &=\log_{60}\frac{60}{5}=\log_{60}12. \end{aligned}$ Therefore the exponent is $\frac12\log_{12}4,$ and $12^{(\frac{1}{2})\log_{12}4}=\sqrt4=2.$

Therefore, the correct answer is B.

26.

The probability that event $A$ occurs is $\frac34;$ the probability that event $B$ occurs is $\frac23.$ Let $p$ be the probability that both $A$ and $B$ occur. The smallest interval necessarily containing $p$ is the interval

$[\frac1{12},\frac12]$

$[\frac5{12},\frac12]$

$[\frac12,\frac23]$

$[\frac5{12},\frac23]$

$[\frac1{12},\frac23]$

Concepts:inclusion-exclusion inequality

Difficulty rating: 2050

Small Hint:

Apply inclusion-exclusion to events $A$ and $B$

Big Hint:

Bound $P(A\cup B)$ between $\max(P(A),P(B))$ and $1$

Solution:

Inclusion-exclusion gives $\begin{aligned} p&=P(A\cap B)\\ &=\frac34+\frac23-P(A\cup B). \end{aligned}$ Since $\frac{3}{4}\le P(A\cup B)\le1,$ it follows that $\frac5{12}\le p\le\frac23.$ Both endpoints can occur, so this is the smallest necessary interval.

Therefore, the correct answer is D.

27.

A large sphere is on a horizontal field on a sunny day. At a certain time the shadow of the sphere reaches out a distance of $10$ m from the point where the sphere touches the ground. At the same instant a meter stick (held vertically with one end on the ground) casts a shadow of length $2$ m. What is the radius of the sphere in meters? (Assume the sun’s rays are parallel and the meter stick is a line segment.)

$\frac52$

$9-4\sqrt5$

$8\sqrt{10}-23$

$6-\sqrt{15}$

$10\sqrt5-20$

Concepts:sphere tangent line distance formula similarity

Difficulty rating: 2310

Small Hint:

In a vertical cross-section, the limiting sun ray is tangent to a circle

Big Hint:

The meter stick shows that the ray rises $1$ unit for every $2$ horizontal units

Solution:

Take the sphere’s ground-contact point as $(0,0)$ and its center as $(0,r).$ The limiting sun ray passes through the shadow endpoint $(10,0)$ and has slope $-\frac{1}{2},$ so its equation is $x+2y-10=0.$ Tangency means the distance from $(0,r)$ to this line equals $r:$ $\frac{10-2r}{\sqrt5}=r.$ Hence $r=\frac{10}{2+\sqrt5}=10\sqrt5-20.$

Therefore, the correct answer is E.

28.

Triangle $ABC$ in the figure has area $10.$ Points $D,$ $E$ and $F,$ all distinct from $A,$ $B$ and $C,$ are on sides $AB,$ $BC$ and $CA$ respectively, and $AD=2,$ $DB=3.$ If triangle $ABE$ and quadrilateral $DBEF$ have equal areas, then that area is

$4$

$5$

$6$

$\frac53\sqrt{10}$

not uniquely determined

Concepts:triangle area area decomposition similarity

Difficulty rating: 2340

Small Hint:

Draw $DE$ and cancel the common triangle $DBE$ from the equal areas

Big Hint:

Equal areas of $ADE$ and $FDE$ imply $AF\parallel DE$ ; then compare triangles $DBE$ and $ABC$

Solution:

Draw $DE.$ From $\begin{aligned} [ABE]&=[ADE]+[DBE],\\ [DBEF]&=[FDE]+[DBE], \end{aligned}$ equality implies $[ADE]=[FDE].$ Their common base $DE$ then gives $AF\parallel DE,$ so triangles $DBE$ and $ABC$ are similar. Hence $\frac{BE}{BC}=\frac{DB}{AB}=\frac{3}{5}.$ Triangles $ABE$ and $ABC$ share the altitude from $A$ to $BC,$ so $[ABE]=(\frac{3}{5})[ABC]=6.$

Therefore, the correct answer is C.

29.

A point $P$ lies in the same plane as a given square of side $1.$ Let the vertices of the square, taken counterclockwise, be $A,$ $B,$ $C$ and $D.$ Also, let the distances from $P$ to $A,$ $B$ and $C,$ respectively, be $u,$ $v$ and $w.$ What is the greatest distance that $P$ can be from $D$ if $u^2+v^2=w^2?$

$1+\sqrt2$

$2\sqrt2$

$2+\sqrt2$

$3\sqrt2$

$3+\sqrt2$

Concepts:coordinate geometry circle distance formula optimization

Difficulty rating: 2310

Small Hint:

Place $D=(0,0),$ $A=(1,0),$ $B=(1,1),$ and $C=(0,1)$

Big Hint:

Substitute $P=(x,y)$ into the distance equation and complete the square

Solution:

Place $D=(0,0),$ $A=(1,0),$ $B=(1,1),$ and $C=(0,1),$ with $P=(x,y).$ Then $\begin{aligned} &(x-1)^2+y^2\\ &\quad+(x-1)^2+(y-1)^2\\ &=x^2+(y-1)^2, \end{aligned}$ which simplifies to $(x-2)^2+y^2=2.$ Thus $P$ lies on a circle centered $2$ units from $D$ with radius $\sqrt2.$ Its greatest distance from $D$ is $2+\sqrt2.$

Therefore, the correct answer is C.

30.

Distinct points $A$ and $B$ are on a semicircle with diameter $MN$ and center $C.$ The point $P$ is on $CN$ and $\angle CAP=\angle CBP=10^\circ.$ If $\overset{\frown}{MA}=40^\circ,$ then $\overset{\frown}{BN}$ equals

$10^\circ$

$15^\circ$

$20^\circ$

$25^\circ$

$30^\circ$

Concepts:circle arc law of sines angle chasing

Difficulty rating: 2430

Small Hint:

The $40^\circ$ arc gives $\angle ACP=140^\circ$

Big Hint:

Compare triangles $ACP$ and $BCP$ using $AC=BC$ and the law of sines

Solution:

Since $\overset{\frown}{MA}=40^\circ,$ the central angle $\angle MCA=40^\circ,$ so $\angle ACP=140^\circ.$ Hence $\angle APC=180^\circ-140^\circ$ ${}-10^\circ=30^\circ.$ Applying the law of sines in triangles $ACP$ and $BCP,$ and using $AC=BC,$ gives $\sin\angle BPC=\sin30^\circ.$ Because $A$ and $B$ are distinct, $\angle BPC=150^\circ.$ Therefore $\begin{aligned} \angle BCN &=180^\circ-150^\circ-10^\circ\\ &=20^\circ, \end{aligned}$ so $\overset{\frown}{BN}=20^\circ.$

Therefore, the correct answer is C.