JEE Mathematics Properties of Triangles MCQs Set 04

Answer: (a) \( \sin^3\theta \)

Explanation:
1. From \( \cot\theta-\cot A=\cot B+\cot C \): \( \frac{\sin(A-\theta)}{\sin\theta\sin A}=\frac{\sin(B+C)}{\sin B\sin C}=\frac{\sin A}{\sin B\sin C} \).
2. Hence \( \sin(A-\theta)=\frac{\sin^2A\sin\theta}{\sin B\sin C} \). Similarly \( \sin(B-\theta)=\frac{\sin^2B\sin\theta}{\sin C\sin A} \) and \( \sin(C-\theta)=\frac{\sin^2C\sin\theta}{\sin A\sin B} \).
3. Multiplying the three, the product is \( \frac{(\sin A\sin B\sin C)^2\sin^3\theta}{(\sin A\sin B\sin C)^2}=\sin^3\theta \).

Teacher's Note:
1. Use \( A+B+C=\pi \) to replace \( \sin(B+C) \) by \( \sin A \).
2. The three expressions are cyclic, so the product collapses neatly.

Answer: (c) \( \frac32 \)

Explanation:
1. Use \( \cos A+\cos B+\cos C=1+4\sin\frac A2\sin\frac B2\sin\frac C2 \).
2. Equivalently, \( \cos A+\cos B+\cos C\le2\sin\frac C2\left(1-\sin\frac C2\right)+1\le\frac32 \), where the second inequality is \( \frac32-2\left(\sin\frac C2-\frac12\right)^2\le\frac32 \).
3. Equality needs \( A=B \) and \( \sin\frac C2=\frac12 \), that is \( A=B=C=\frac\pi3 \). Check: \( 3\cos\frac\pi3=\frac32 \).

Teacher's Note:
1. The maximum occurs for the equilateral triangle.
2. Completing the square in \( \sin\frac C2 \) shows the bound neatly.

Answer: (b) \( P\in[3,4) \)

Explanation:
1. \( (a-b)^2+(b-c)^2+(c-a)^2\ge0 \Rightarrow a^2+b^2+c^2\ge ab+bc+ca \), so \( (a+b+c)^2\ge3(ab+bc+ca) \) and \( P\ge3 \).
2. In a triangle \( a>|b-c| \), etc. Squaring and adding gives \( a^2+b^2+c^2<2(ab+bc+ca) \), so \( (a+b+c)^2<4(ab+bc+ca) \) and \( P<4 \).
3. Hence \( 3\le P<4 \), i.e. \( P\in[3,4) \).

Teacher's Note:
1. Two standard bounds: one from squares being non-negative, one from the triangle inequality.
2. Equality in the lower bound holds for the equilateral triangle.

Answer: (b) \( \tan\left(\frac C2-\frac\pi4\right) \)

Explanation:
1. By the sine rule and \( \sin^2A-\sin^2B=\sin(A+B)\sin(A-B) \): \( \frac{\sin^2A+\sin^2B}{\sin(A+B)\sin(A-B)}\sin(A-B)=1 \), so \( \sin^2A+\sin^2B=\sin C \).
2. Then \( 2-\cos2A-\cos2B=2\sin C \), i.e. \( \cos2A+\cos2B=2(1-\sin C) \), and \( 2\cos(A+B)\cos(A-B)=2(1-\sin C) \).
3. Since \( \cos(A+B)=-\cos C \): \( \cos(A-B)=\frac{1-\sin C}{-\cos C}=\frac{\left(\sin\frac C2-\cos\frac C2\right)^2}{\sin^2\frac C2-\cos^2\frac C2}=\frac{\sin\frac C2-\cos\frac C2}{\sin\frac C2+\cos\frac C2}=\tan\left(\frac C2-\frac\pi4\right) \).

Teacher's Note:
1. Convert sides to sines and use \( \sin^2A-\sin^2B=\sin(A+B)\sin(A-B) \).
2. \( 1-\sin C=\left(\sin\frac C2-\cos\frac C2\right)^2 \) is the key simplification.

Answer: (a) \( \frac{41}{9} \)

Explanation:
1. Let \( O \) be the foot of the tower. \( OA=h\cot\alpha=\frac{h}{0.6}=\frac{5h}{3} \) and \( OB=\frac{h}{0.75}=\frac{4h}{3} \).
2. \( A \) is south and \( B \) is east of \( O \), so \( \angle AOB=90^\circ \).
3. \( AB^2=OA^2+OB^2=\frac{25h^2}{9}+\frac{16h^2}{9}=\frac{41}{9}h^2 \), so \( \lambda^2=\frac{41}{9} \).

Teacher's Note:
1. Due south and due east are perpendicular directions, so the Pythagoras theorem applies at the foot of the tower.
2. Distance from the foot is \( h\cot(\text{elevation}) \).

Answer: (c) \( a=h(\sqrt3-1) \)

Explanation:
1. Take \( A \) as the origin and \( D=(h,0),\ C=(h,h) \) (since \( \angle CAD=45^\circ \)). Then \( B=\left(\frac{\sqrt3a}{2},\frac a2\right) \).
2. From \( B \), the horizontal distance to the tower is \( h-\frac{\sqrt3a}{2} \) and the height of \( C \) above \( B \) is \( h-\frac a2 \). For a \( 60^\circ \) elevation: \( h-\frac a2=\sqrt3\left(h-\frac{\sqrt3a}{2}\right) \).
3. So \( \sqrt3h-h=\frac{3a}2-\frac a2=a \), giving \( a=h(\sqrt3-1) \).

Teacher's Note:
1. Coordinates make elevation problems with two non-horizontal lines easy: elevation is the slope of the line of sight.
2. Check: \( a=0.732h \), \( B=(0.634h,0.366h) \) and the slope to \( C \) is \( \frac{0.634h}{0.366h}\approx1.73=\tan60^\circ \).

Answer: (b) (i), (ii) and (iv) only

Explanation:
1. \( b\cos^2\frac A2+a\cos^2\frac B2=\frac{a+b}2+\frac{b\cos A+a\cos B}2=\frac{a+b+c}2 \) (projection formula \( c=a\cos B+b\cos A \)). Setting this equal to \( \frac{3c}2 \) gives \( a+b=2c \).
2. (i) \( c=\frac{a+b}2\ge\sqrt{ab} \), so \( c^2\ge ab \) is true. (ii) \( 2c=a+b\ge2\sqrt{ab}>\sqrt{ab} \) is true.
3. (iii) With \( 2c-a=b \) and \( 2c-b=a \), the expression is \( \frac{a+c}b+\frac{b+c}a \). For \( a=1+t,\ b=1-t,\ c=1 \) it equals \( \frac{4+2t^2}{1-t^2}\ge4 \), so the statement \( \le4 \) fails except at \( t=0 \).
4. (iv) By AM-GM, \( \frac ac+\frac cb+\frac ba\ge3\sqrt[3]{\frac ac\cdot\frac cb\cdot\frac ba}=3 \), which is true.

Teacher's Note:
1. The projection formula \( c=a\cos B+b\cos A \) is the fastest route to simplify such conditions.
2. Test inequalities with a simple example such as \( a=1.2,\ b=0.8,\ c=1 \) before accepting them.

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. Let the sides be \( a,ar,ar^2 \). The triangle inequality \( a+ar>ar^2 \) gives \( r^2-r-1<0 \), so \( 1<r<\frac{1+\sqrt5}2 \). (i) is true.
2. Since \( r>1 \), \( a<b<c \) and so \( A<B<C \), giving \( C>\frac\pi3 \) (the largest angle exceeds the average). (iv) is true.
3. \( \cos B=\frac{a^2+a^2r^4-a^2r^2}{2a^2r^2}=\frac12\left[r^2+\frac1{r^2}-1\right]=\frac12\left[\left(r-\frac1r\right)^2+1\right]>\frac12 \), so \( B<\frac\pi3 \). Hence (ii) is true and (iii) is false.

Teacher's Note:
1. The largest side faces the largest angle, and the angles add to \( \pi \).
2. \( \cos B>\frac12 \) means \( B<\frac\pi3 \) because cosine decreases on \( (0,\pi) \).

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. Tangent lengths from a vertex are equal. Let \( BP=BR=n,\ CQ=CP=n+1,\ AR=AQ=n+2 \). Then \( BC=2n+1,\ CA=2n+3,\ AB=2n+2 \) and \( s=3n+3 \).
2. The inradius satisfies \( r^2=\frac{(s-a)(s-b)(s-c)}s=\frac{(n+2)\,n\,(n+1)}{3n+3}=\frac{n(n+2)}3=16 \), so \( n(n+2)=48 \) and \( n=6 \).
3. The sides are \( 13,14,15 \): consecutive integers and in A.P. (i), (ii) true. The perimeter is \( 42 \) (iii) true.
4. \( \Delta=\sqrt{21\cdot8\cdot7\cdot6}=84 \) and \( R=\frac{abc}{4\Delta}=\frac{2730}{336}=\frac{65}8 \), so the circum-diameter is \( \frac{65}4 \), not 65. (iv) is false.

Teacher's Note:
1. Tangents from one point to a circle are equal, which converts the problem to three unknown lengths.
2. Always compute the final quantity (here the circumdiameter) rather than trusting a partial match.

Answer: (b) (ii) and (iii) only

Explanation:
1. \( \cos A,\cos B,\cos C \) in A.P. means \( 2\cos B=\cos A+\cos C \). Write \( \cos A+\cos C=2\sin\frac B2\cos\frac{A-C}2 \) and \( 2\cos B=2\left(\cos^2\frac B2-\sin^2\frac B2\right) \). Then \( \cos^2\frac B2=\sin\frac B2\left[\cos\frac{A+C}2+\cos\frac{A-C}2\right]=2\sin\frac B2\cos\frac A2\cos\frac C2 \).
2. Dividing, \( \cot\frac B2=\frac{2\cos\frac A2\cos\frac C2}{\sin\frac{A+C}2}=\frac2{\tan\frac A2+\tan\frac C2} \), so \( \tan\frac A2+\tan\frac C2=2\tan\frac B2 \): the half-angle tangents are in A.P.
3. Hence \( \cot\frac A2,\cot\frac B2,\cot\frac C2 \) are in H.P. (ii true, iv false).
4. Since \( r_1=s\tan\frac A2 \) etc., \( r_1,r_2,r_3 \) are in A.P. (iii true). Nothing forces \( a,b,c \) to be in H.P. (i false).

Teacher's Note:
1. Use \( r_1=s\tan\frac A2 \), so exradii are proportional to the half-angle tangents.
2. In a triangle, A.P. in cosines is a different condition from A.P. in the sides.

Answer: (a) 6

Explanation:
1. From \( \frac12ah_1=\Delta=rs \): \( \frac{h_1}r=\frac{2s}a=\frac{a+b+c}a \), so \( \frac{h_1+r}{h_1-r}=\frac{2a+b+c}{b+c}=\frac{2(a+b+c)}{b+c}-1 \).
2. The sum is \( 2(a+b+c)\left[\frac1{b+c}+\frac1{c+a}+\frac1{a+b}\right]-3 \).
3. By AM-HM, \( \frac1{b+c}+\frac1{c+a}+\frac1{a+b}\ge\frac9{2(a+b+c)} \), so the sum is at least \( 9-3=6 \), attained for the equilateral triangle.

Teacher's Note:
1. Express each altitude through the sides using \( h_1=\frac{2\Delta}a \) and \( r=\frac\Delta s \).
2. Check the equilateral case: each term equals 2, so the total is 6.

Answer: (a) 3

Explanation:
1. The cotangents \( \cot\frac A2,\cot\frac B2,\cot\frac C2 \) are in A.P., so \( \cot\frac A2+\cot\frac C2=2\cot\frac B2 \).
2. In any triangle \( \cot\frac A2+\cot\frac B2+\cot\frac C2=\cot\frac A2\cot\frac B2\cot\frac C2 \). Using the A.P. condition, \( 3\cot\frac B2=\cot\frac B2\cdot\cot\frac A2\cot\frac C2 \), so \( \cot\frac A2\cot\frac C2=3 \).
3. By AM-GM, \( \frac{\cot\frac A2+\cot\frac C2}2\ge\sqrt3 \), that is \( \cot\frac B2\ge\sqrt3 \), so \( \cot^2\frac B2\ge3 \).

Teacher's Note:
1. The identity \( \sum\cot\frac A2=\prod\cot\frac A2 \) follows from \( A+B+C=\pi \).
2. Use AM-GM on the two end terms of the A.P.

Answer: (a) 5

Explanation:
1. The altitudes of \( \triangle A'B'C' \) are \( \frac{2\Delta'}{h_a}=\frac{2\Delta'}{2\Delta/a}=\frac{a\Delta'}\Delta \), and similarly for the other sides.
2. So \( \triangle A''B''C'' \) is similar to \( \triangle ABC \) with ratio \( \frac{\Delta'}\Delta \), and \( \Delta''=\Delta\left(\frac{\Delta'}\Delta\right)^2=\frac{\Delta'^2}\Delta \).
3. \( \Delta=\frac{\Delta'^2}{\Delta''}=\frac{900}{20}=45 \), so \( \frac\Delta9=5 \).

Teacher's Note:
1. Areas of similar triangles scale as the square of the ratio of sides.
2. The altitude triangle of the altitude triangle is similar to the original.

Answer: (a) 6

Explanation:
1. The incircle has radius \( r=\frac23 \). With \( p_1=\frac{2\Delta}a \) etc., by AM-GM \( p_1+p_2+p_3\ge3(p_1p_2p_3)^{1/3}=6\Delta\left(\frac1{abc}\right)^{1/3} \).
2. With \( \Delta=rs \): \( =6rs\left(\frac1{abc}\right)^{1/3}=9r\left(\frac{a+b+c}3\right)\left(\frac1{abc}\right)^{1/3}\ge9r \) (AM-GM on \( a,b,c \)).
3. Equality holds for the equilateral triangle, so the minimum is \( 9r=9\times\frac23=6 \).

Teacher's Note:
1. Rewrite altitudes as \( 2rs/a \) so the sum is \( 2rs\left(\frac1a+\frac1b+\frac1c\right) \) and use AM-HM.
2. The minimum is for the equilateral triangle.

Answer: (c) \( x^3-x^2(4R+r)+s^2x-rs^2=0 \)

Explanation:
1. \( r_1+r_2+r_3=\Delta\left[\frac1{s-a}+\frac1{s-b}+\frac1{s-c}\right]=4R+r \).
2. \( r_1r_2+r_2r_3+r_3r_1=\Delta^2\left[\frac1{(s-a)(s-b)}+\dots\right]=\Delta^2\frac{s}{(s-a)(s-b)(s-c)}=s^2 \).
3. \( r_1r_2r_3=\frac{\Delta^3}{(s-a)(s-b)(s-c)}=\Delta s=rs^2 \). The cubic is \( x^3-(4R+r)x^2+s^2x-rs^2=0 \).

Teacher's Note:
1. Memorise the symmetric functions of the exradii: \( 4R+r \), \( s^2 \) and \( rs^2 \).
2. Use \( \Delta^2=s(s-a)(s-b)(s-c) \) to simplify.

Answer: (a) \( 2s^2(s+r+2R) \)

Explanation:
1. The exradii are the roots of \( x^3-(4R+r)x^2+s^2x-rs^2=(x-r_1)(x-r_2)(x-r_3) \).
2. Put \( x=-s \): \( -s^3-(4R+r)s^2-s^3-rs^2=(-s-r_1)(-s-r_2)(-s-r_3)=-(s+r_1)(s+r_2)(s+r_3) \).
3. So the product is \( 2s^3+(4R+2r)s^2=2s^2(s+r+2R) \).

Teacher's Note:
1. Evaluating a polynomial at \( x=-s \) gives the product of \( (-s-r_i) \), a sign change from the required product.
2. Numerical check for any triangle supports the identity.

Answer: (b) \( a>b\sin A \) and \( a<b \)

Explanation:
1. For two different triangles the quadratic in \( c \) must have two distinct positive roots.
2. Distinct real roots need \( D>0 \), i.e. \( a>b\sin A \). Both positive needs sum \( 2b\cos A>0 \) (A acute) and product \( b^2-a^2>0 \), i.e. \( a<b \).
3. So the condition is \( a>b\sin A \) and \( a<b \) (with A acute, which follows from \( a<b \)).

Teacher's Note:
1. Both roots must be positive, so check sum and product of roots.
2. The ambiguous case occurs when the side opposite the given angle lies between \( b\sin A \) and \( b \).

Answer: (b) only one triangle is possible

Explanation:
1. With \( a=b \), \( \sqrt{a^2-b^2\sin^2A}=\sqrt{b^2\cos^2A}=b\cos A \) (A is acute, since \( A=B \)).
2. The roots are \( c=b\cos A\pm b\cos A \), that is \( c=2b\cos A>0 \) and \( c=0 \).
3. The root \( c=0 \) is rejected, so only one triangle is possible.

Teacher's Note:
1. A side of length zero does not give a triangle.
2. In the isosceles case \( a=b \) the ambiguity disappears.

Answer: (a) \( \frac34 \)

Explanation:
1. \( \frac r{r_1}=\frac{4R\sin\frac A2\sin\frac B2\sin\frac C2}{4R\sin\frac A2\cos\frac B2\cos\frac C2}=\tan\frac B2\tan\frac C2=\frac14 \).
2. Using \( \tan\frac A2\tan\frac B2+\tan\frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1 \), \( \tan\frac A2\left(\tan\frac B2+\tan\frac C2\right)=1-\tan\frac B2\tan\frac C2 \).
3. That equals \( 1-\frac14=\frac34 \).

Teacher's Note:
1. Use \( r=4R\prod\sin\frac A2 \) and \( r_1=4R\sin\frac A2\cos\frac B2\cos\frac C2 \).
2. The identity \( \sum\tan\frac A2\tan\frac B2=1 \) holds in every triangle.

Answer: (a) 27

Explanation:
1. \( r_1r_2r_3=\frac{\Delta^3}{(s-a)(s-b)(s-c)} \) and \( r^3=\frac{\Delta^3}{s^3} \), so the ratio is \( \frac{s^3}{(s-a)(s-b)(s-c)} \).
2. By AM-GM, \( (s-a)(s-b)(s-c)\le\left(\frac{(s-a)+(s-b)+(s-c)}3\right)^3=\frac{s^3}{27} \).
3. So the ratio is at least 27, with equality for the equilateral triangle.

Teacher's Note:
1. Express everything in terms of \( s-a,s-b,s-c \), which add up to \( s \).
2. Equality in AM-GM means \( a=b=c \).

Answer: (a) \( \frac34 \)

Explanation:
1. With \( a=2b-c \) the triangle inequalities become: \( a+b>c\Rightarrow3b>2c\Rightarrow\frac bc>\frac23 \); \( b+c>a\Rightarrow2c>b\Rightarrow\frac bc<2 \); and \( c+a>b\Rightarrow2b>b \), always true.
2. So \( \frac bc\in\left(\frac23,2\right) \).
3. Among the options only \( \frac34 \) lies in this interval (\( \frac23<\frac34<2 \)).

Teacher's Note:
1. Write the third side in terms of the other two and apply all three triangle inequalities.
2. Test each option against the open interval.

Answer: (a) 1

Explanation:
1. \( \frac{PD}{AD}=\frac{\text{ar}(\triangle PBC)}{\text{ar}(\triangle ABC)} \), since triangles with the same base \( BC \) have areas in the ratio of their heights, which are in the ratio \( PD:AD \).
2. Similarly \( \frac{PE}{BE}=\frac{\text{ar}(\triangle PCA)}{\text{ar}(\triangle ABC)} \) and \( \frac{PF}{CF}=\frac{\text{ar}(\triangle PAB)}{\text{ar}(\triangle ABC)} \).
3. The three areas add up to \( \text{ar}(\triangle ABC) \), so the sum is 1.

Teacher's Note:
1. Ratios of segments along a cevian equal ratios of areas of triangles on a common base.
2. The three small triangles tile the whole triangle.

Answer: (a) \( \frac34 \)

Explanation:
1. \( \alpha^2=\frac14(2b^2+2c^2-a^2),\ \beta^2=\frac14(2c^2+2a^2-b^2),\ \gamma^2=\frac14(2a^2+2b^2-c^2) \).
2. Adding: \( \alpha^2+\beta^2+\gamma^2=\frac14(3a^2+3b^2+3c^2) \).
3. The ratio is \( \frac34 \).

Teacher's Note:
1. Apollonius' theorem gives each median length directly.
2. The sum of squares of medians is always \( \frac34 \) of the sum of squares of sides.

Answer: (a) \( \sqrt3 \)

Explanation:
1. Join \( P \) to the vertices. The area of \( \triangle ABC \) is the sum of the areas of \( \triangle PBC,\triangle PCA,\triangle PAB \).
2. \( \frac{\sqrt3}4\times2^2=\frac12(2x+2y+2z) \), i.e. \( \sqrt3=x+y+z \).

Teacher's Note:
1. The sum of the perpendicular distances from an interior point to the sides of an equilateral triangle equals its altitude.
2. The altitude here is \( \frac{\sqrt3}2\times2=\sqrt3 \).

Answer: (a) 1

Explanation:
1. Angles in A.P.: \( 2B=A+C \Rightarrow B=\frac\pi3,\ A+C=\frac{2\pi}3 \). Sides in G.P.: \( b^2=ac \Rightarrow\sin^2B=\sin A\sin C \), so \( \sin A\sin C=\frac34 \).
2. \( \sin A\sin C=\frac12[\cos(A-C)-\cos(A+C)]=\frac12\left[\cos(A-C)+\frac12\right]=\frac34 \), giving \( \cos(A-C)=1 \), so \( A=C \).
3. Hence \( A=B=C=\frac\pi3 \): the triangle is equilateral, \( a=b=c \), and the ratio is \( \frac{3a^3}{3a^3}=1 \).

Teacher's Note:
1. Combining an A.P. of angles with a G.P. of sides forces an equilateral triangle.
2. Once equilateral, substitute \( a=b=c \).

Answer: (a) 4

Explanation:
1. \( abc=4R\Delta \) and \( a+b+c=2s \), so \( \frac{abc(a+b+c)}{\Delta^2}=\frac{8Rs}{\Delta}=\frac{8R}r \) (using \( \Delta=rs \)).
2. Euler's inequality gives \( R\ge2r \), so \( \frac{8R}r\ge16 \).
3. Therefore \( \frac{\sqrt{abc(a+b+c)}}\Delta\ge4 \), with equality for the equilateral triangle.

Teacher's Note:
1. Rewrite every term using \( R \), \( r \), \( s \) and \( \Delta \).
2. Euler's inequality \( R\ge2r \) is the key inequality here.