JEE Mathematics Properties of Triangles MCQs Set 02

Answer: (b) 23

Explanation:
1. Cosine rule: \( c^2=17^2+19^2-2\cdot17\cdot19\cos\alpha=650-646\cos\alpha \).
2. As \( \alpha \) goes from 0 to \( \tfrac\pi2 \), \( \cos\alpha \) decreases from 1 to 0, so \( c^2\in(4,650) \) and \( c\in(2,\ 25.49\ldots) \).
3. Integers from 3 to 25: 23 values.

Teacher's Note:
1. Use the cosine rule and let the angle sweep over its allowed interval.
2. The end values 2 and 25.49 are not attained because the angle is strictly between 0 and 90 degrees.

Answer: (b) r(r+2R)

Explanation:
1. Let the right angle be at C so the hypotenuse \( c=2R \). For a right triangle \( r=\tfrac{a+b-c}2 \), so \( a+b=2r+2R \).
2. From \( a^2+b^2=4R^2 \): \( 2ab=(a+b)^2-4R^2=4r^2+8rR \), so \( ab=2r^2+4rR \).
3. Area \( =\tfrac{ab}2=r^2+2rR=r(r+2R) \).

Teacher's Note:
1. In a right triangle the hypotenuse is a diameter of the circumcircle and r=(a+b-c)/2.
2. Use (a+b)^2 - (a^2+b^2) = 2ab.

Answer: (c) \( 90^\circ,45^\circ,45^\circ \)

Explanation:
1. Since \( a=2R\sin A \), the given relation gives \( \sin A=\frac{b+c}{2\sqrt{bc}} \).
2. By A.M.-G.M. \( \tfrac{b+c}{2\sqrt{bc}}\ge1 \), while \( \sin A\le1 \). So both equal 1: \( A=90^\circ \) and \( b=c \).
3. The triangle is right-angled and isosceles: \( 90^\circ,45^\circ,45^\circ \).

Teacher's Note:
1. Replace R by a/(2 sin A) and compare with A.M.-G.M.
2. An inequality squeezed from both sides forces equality.

Answer: (b) 100

Explanation:
1. Let \( A=\alpha \), so \( B=C=90^\circ-\tfrac\alpha2 \) and \( \angle DBC=45^\circ-\tfrac\alpha4 \). Apply the sine rule in triangles ABD and BDC and impose \( BC=BD+AD \); the equation is satisfied exactly at \( \alpha=100^\circ \).
2. Check with \( A=100^\circ,\ B=C=40^\circ \): in triangle BDC, \( BD=\dfrac{BC\sin40^\circ}{\sin120^\circ} \); in triangle ABD, \( AD=\dfrac{AB\sin20^\circ}{\sin60^\circ} \) with \( AB=\dfrac{BC\sin40^\circ}{\sin100^\circ} \).
3. Numerically \( BD+AD=1.000\times BC \), as required. Hence \( A=100^\circ \).

Teacher's Note:
1. The condition BC = BD + AD fixes the apex angle; the sine rule in the two small triangles turns it into a trigonometric equation.
2. Always verify a candidate angle by substituting back.

Answer: (c) \( \tfrac14 \)

Explanation:
1. Known results: \( r_1r_2+r_2r_3+r_3r_1=s^2 \) and \( r_1+r_2=4R\cos^2\tfrac C2 \) etc.
2. Using \( r_1+r_2=4R\cos^2\tfrac C2 \) (and cyclic) together with \( \prod\cos\tfrac A2=\tfrac{s}{4R} \): \( (r_1+r_2)(r_2+r_3)(r_3+r_1)=64R^3\cdot\tfrac{s^2}{16R^2}=4Rs^2 \).
3. Therefore the fraction equals \( \dfrac{4Rs^2}{s^2}=4R \), so \( R=\tfrac14\cdot4R \) and \( k=\tfrac14 \).

Teacher's Note:
1. Express each exradius sum through the half-angle cosines.
2. The product of the half-angle cosines equals s/(4R).

Answer: (c) \( \tfrac12Rr^2 \)

Explanation:
1. In triangle MIN, \( IM=IN=r \) and \( \angle MIN=180^\circ-A \), so \( MN=2r\cos\tfrac A2 \).
2. Circumradius \( x=\frac{MN}{2\sin(180^\circ-A)}=\frac{2r\cos\frac A2}{2\sin A}=\frac{r}{2\sin\frac A2} \). Similarly for y and z.
3. Product \( xyz=\frac{r^3}{8\prod\sin\frac A2} \) and \( \prod\sin\tfrac A2=\tfrac r{4R} \), so \( xyz=\tfrac{r^3\cdot4R}{8r}=\tfrac12Rr^2 \).

Teacher's Note:
1. The angle at the incentre in the triangle formed by two touch points is 180 degrees minus the vertex angle.
2. Use r = 4R times the product of half-angle sines.

Answer: (c) \( \tfrac rR \)

Explanation:
1. The pedal (orthic) triangle has sides \( a\cos A,\ b\cos B,\ c\cos C \) and perimeter \( R(\sin2A+\sin2B+\sin2C)=4R\sin A\sin B\sin C \) (acute triangle).
2. Since \( \Delta=2R^2\sin A\sin B\sin C \), the perimeter is \( \dfrac{2\Delta}R \).
3. Ratio to the perimeter \( 2s \): \( \dfrac{2\Delta/R}{2s}=\dfrac{\Delta}{sR}=\dfrac rR \).

Teacher's Note:
1. The side of the orthic triangle opposite A is a cos A.
2. Use Delta = rs to convert.

Answer: (c) 2

Explanation:
1. Use \( \cot\tfrac B2\cot\tfrac C2=\dfrac{s}{s-a} \).
2. With \( b+c=3a \), \( 2s=4a \) so \( s=2a \) and \( s-a=a \).
3. The value is \( \tfrac{2a}{a}=2 \).

Teacher's Note:
1. Half-angle formulas relate cot(B/2)cot(C/2) to s and s-a.
2. Express s through a using the given condition.

Answer: (b) \( \Delta \)

Explanation:
1. \( \dfrac{a^2}{\sin A}=a\cdot2R \) so the bracket equals \( 2R(a+b+c)=4Rs \).
2. Also \( \prod\sin\tfrac A2=\tfrac r{4R} \).
3. Product \( =4Rs\cdot\tfrac r{4R}=rs=\Delta \).

Teacher's Note:
1. Use the sine rule to simplify each ratio to 2R times the side.
2. r s equals the area of the triangle.

Answer: (b) 15

Explanation:
1. Let \( AD=h \), \( \angle BAD=\theta_1 \), \( \angle DAC=\theta_2 \) with \( \tan\theta_1=\tfrac3h,\ \tan\theta_2=\tfrac2h \).
2. Since \( \theta_1+\theta_2=45^\circ \): \( \dfrac{\frac5h}{1-\frac6{h^2}}=1 \), so \( h^2-5h-6=0 \) and \( h=6 \).
3. Area \( =\tfrac12\cdot5\cdot6=15 \).

Teacher's Note:
1. Split the 45 degree angle at A into the two angles made with the altitude.
2. Use the tangent addition formula.

Answer: (d) 40

Explanation:
1. The apex angle is \( 60^\circ \). By the sine rule the other two sides are \( \dfrac{10\sin50^\circ}{\sin60^\circ} \) and \( \dfrac{10\sin70^\circ}{\sin60^\circ} \).
2. Perimeter \( =10+\dfrac{10(\sin50^\circ+\sin70^\circ)}{\sin60^\circ}=10+\dfrac{10\cdot2\sin60^\circ\cos10^\circ}{\sin60^\circ}=10+20\cos10^\circ \).
3. So \( x=10,\ y=20,\ z=10 \) and the sum is 40.

Teacher's Note:
1. Find the third angle first, then apply the sine rule to the two unknown sides.
2. Sum of sines: sin50 + sin70 = 2 sin60 cos10.

Answer: (a) \( \tfrac\pi6 \)

Explanation:
1. Sine rule: \( \dfrac{\sin A}{BC}=\dfrac{\sin B}{AC} \), so \( \sin A=\tfrac12\sin B\le\tfrac12 \).
2. Angle A is opposite the shorter side BC, so it is acute: \( A\le\tfrac\pi6 \), attained when \( B=\tfrac\pi2 \).

Teacher's Note:
1. Angle A is opposite BC and B is opposite AC.
2. The largest sine value occurs when angle B is a right angle.

Answer: (c) \( R'=R \)

Explanation:
1. In triangle OBC, \( BC=a \) and \( \angle BOC=180^\circ-A \) (acute triangle).
2. Circumradius \( R'=\dfrac{a}{2\sin(180^\circ-A)}=\dfrac{a}{2\sin A}=R \).

Teacher's Note:
1. The angle BOC at the orthocentre is 180 degrees minus A.
2. The circle through B, O, C is the reflection of the circumcircle in BC.

Answer: (c) right isosceles

Explanation:
1. With \( a=2R\sin A \): \( b+c=2\sin A\sqrt{bc} \), so \( \sin A=\dfrac{b+c}{2\sqrt{bc}}\ge1 \).
2. This forces \( \sin A=1 \) and \( b=c \): a right isosceles triangle.

Teacher's Note:
1. Substitute a = 2R sin A.
2. A.M.-G.M. gives sin A at least 1, so equality must hold.

Answer: (c) \( \tfrac{30}7 \)

Explanation:
1. Let the shortest side \( a=6 \); \( s=\tfrac{21}2 \). The line through I parallel to BC is at distance r from BC, so it cuts the triangle similar to ABC at distance \( h_a-r \) from A.
2. Ratio \( \frac{PQ}{a}=\frac{h_a-r}{h_a}=1-\frac r{h_a}=1-\frac{a}{2s} \) since \( r=\tfrac\Delta s \) and \( h_a=\tfrac{2\Delta}a \).
3. So \( PQ=6\left(1-\tfrac{6}{21}\right)=\tfrac{30}7 \).

Teacher's Note:
1. PQ is parallel to BC, so triangle APQ is similar to triangle ABC.
2. The ratio of similarity is the ratio of the distances from A.

Answer: (c) \( \sqrt{45} \)

Explanation:
1. Let \( BP=PQ=QC=x \), so \( BC=3x \). The mid-point M of BC is also the mid-point of PQ, and \( AM=\tfrac{BC}2=\tfrac{3x}2 \) (median to the hypotenuse).
2. In triangle APQ the median \( AM^2=\dfrac{2AP^2+2AQ^2-PQ^2}{4}=\dfrac{18+32-x^2}{4} \).
3. So \( \tfrac{9x^2}4=\tfrac{50-x^2}4 \), \( x^2=5 \) and \( BC^2=9x^2=45 \).

Teacher's Note:
1. The midpoint of the hypotenuse is equidistant from all three vertices.
2. Apply the median length formula in triangle APQ.

Answer: (b) \( 2\Delta \)

Explanation:
1. A.M. of sides \( =\tfrac{a+b+c}3=\tfrac{2s}3 \).
2. Since \( \tfrac1{h_a}=\tfrac a{2\Delta} \): \( \sum\tfrac1{h}=\tfrac{2s}{2\Delta}=\tfrac s\Delta \), so H.M. of altitudes \( =\tfrac{3}{s/\Delta}=\tfrac{3\Delta}s \).
3. Product \( =\tfrac{2s}3\cdot\tfrac{3\Delta}s=2\Delta \).

Teacher's Note:
1. Altitude to side a is 2 Delta divided by a.
2. The harmonic mean of three numbers is 3 divided by the sum of reciprocals.

Answer: (b) (9, -3)

Explanation:
1. Put \( B=(0,0),C=(12,0),A=(6,6\sqrt3) \) and the mid-point \( M=(6,0) \). Let the side be s and let MP make angle \( \alpha \) with MB, with P on AB and Q on AC.
2. P on line AB (\( y=\sqrt3x \)): \( s\sin\alpha=\sqrt3(6-s\cos\alpha) \), giving \( s=\dfrac{3\sqrt3}{\sin(\alpha+60^\circ)} \). Q on AC (with MQ perpendicular to MP): \( s=\dfrac{3\sqrt3}{\sin(\alpha+30^\circ)} \).
3. Equal sines give \( \alpha+60^\circ=180^\circ-(\alpha+30^\circ) \), so \( \alpha=45^\circ \).
4. Then \( s=\dfrac{3\sqrt3}{\sin105^\circ}=3\sqrt3(\sqrt6-\sqrt2)=9\sqrt2-3\sqrt6 \), so \( (p,q)=(9,-3) \).

Teacher's Note:
1. Use coordinates with BC on the x-axis and set the two adjacent vertices on the slanted sides.
2. MQ is MP rotated by 90 degrees, which gives two linear conditions in s.

Answer: (a) \( \dfrac{2\Delta}{a+b} \)

Explanation:
1. Let the centre O lie on AB and the radius be \( \rho \). The semicircle touches AC and BC, so the perpendiculars from O to AC and BC both equal \( \rho \).
2. Split the triangle into triangles AOC and BOC: \( \Delta=\tfrac12b\rho+\tfrac12a\rho \).
3. So \( \rho=\dfrac{2\Delta}{a+b} \).

Teacher's Note:
1. Join O to C and add the areas of the two triangles on either side.
2. Each has height equal to the radius.

Answer: (c) equilateral

Explanation:
1. The side BC subtends \( \angle BHC=180^\circ-A \) at the orthocentre H and \( \angle BIC=90^\circ+\tfrac A2 \) at the incentre I.
2. Equal: \( 180^\circ-A=90^\circ+\tfrac A2 \), so \( A=60^\circ \). The same holds for each side, so \( A=B=C=60^\circ \).

Teacher's Note:
1. Write the two angles subtended by a side in terms of the opposite vertex angle.
2. Apply the condition to every side, not only one.

Answer: (b) B and C only

Explanation:
1. \( \cos A+\cos B=2\sin\tfrac C2\cos\tfrac{A-B}2 \), so \( \cos\tfrac{A-B}2=2\sin\tfrac C2=2\cos\tfrac{A+B}2 \).
2. Expand: \( \cos\tfrac A2\cos\tfrac B2+\sin\tfrac A2\sin\tfrac B2=2\cos\tfrac A2\cos\tfrac B2-2\sin\tfrac A2\sin\tfrac B2 \), so \( \cot\tfrac A2\cot\tfrac B2=3 \).
3. Since \( \cot\tfrac A2\cot\tfrac B2=\tfrac s{s-c} \): \( s=3s-3c \), \( a+b=2c \), so a, c, b are in A.P. Statements B and C are true.

Teacher's Note:
1. Convert the sum of cosines to a product, then equate it to 4 sin^2(C/2).
2. cot(A/2)cot(B/2) equals s/(s-c).

Answer: (c) A, B, C and D all are correct

Explanation:
1. \( \Delta=rs=5 \) (A true).
2. \( abc=4R\Delta=60 \) (B true).
3. \( a^2+b^2+c^2=2(s^2-r^2-4Rr)=2(25-1-12)=24 \) (C true).
4. \( r_1+r_2+r_3=4R+r=13 \) (D true).

Teacher's Note:
1. Use the standard identities Delta = rs, abc = 4R Delta, and r1+r2+r3 = 4R + r.
2. a^2+b^2+c^2 = 2(s^2 - r^2 - 4Rr).

Answer: (b) A, B and D only

Explanation:
1. \( r\rho=\dfrac{\Delta}{s}\cdot\dfrac{\Delta}{s-a}=(s-b)(s-c) \).
2. For \( b=c \): \( s-b=\tfrac a2 \), so \( r\rho=\tfrac{a^2}4 \) (D).
3. Also \( a=2R\sin A \) gives \( \tfrac{a^2}4=R^2\sin^2A \) (A), and \( A=180^\circ-2B \) gives \( \sin A=\sin2B \) so (B) holds. (C) is false.

Teacher's Note:
1. r times r_a simplifies to (s-b)(s-c).
2. In an isosceles triangle with b = c, s-b equals a/2.

Answer: (b) \( 60^\circ \) or \( 120^\circ \)

Explanation:
1. The cosine rule \( c^2-2bc\cos A+b^2-a^2=0 \) has roots \( c_1,c_2 \): \( c_1+c_2=2b\cos A \) and \( c_1c_2=b^2-a^2 \).
2. The condition: \( (c_1+c_2)^2-c_1c_2=a^2 \) gives \( 4b^2\cos^2A-b^2+a^2=a^2 \), so \( \cos^2A=\tfrac14 \).
3. Hence \( A=60^\circ \) or \( 120^\circ \).

Teacher's Note:
1. Treat the cosine rule as a quadratic in the unknown side c.
2. Use the sum and product of its two roots.

Answer: (b) 7

Explanation:
1. Let \( AB=AC=b,\ BC=a \). The medians from B and C are equal, \( m^2=\dfrac{2(a^2+b^2)-b^2}{4}=\dfrac{2a^2+b^2}4 \).
2. Let G be the centroid: \( GB=GC=\tfrac23m \). Perpendicular medians give \( GB^2+GC^2=a^2 \), so \( \tfrac89m^2=a^2 \).
3. Substitute: \( \tfrac29(2a^2+b^2)=a^2 \), so \( b^2=\tfrac52a^2 \). Thus \( \tfrac pq=\tfrac52 \) and \( p+q=7 \).

Teacher's Note:
1. The centroid divides each median in the ratio 2:1.
2. Right angle at G gives the Pythagoras relation GB^2 + GC^2 = BC^2.

Answer: (a) 0

Explanation:
1. \( r_1r_2r_3=\dfrac{\Delta^3}{(s-a)(s-b)(s-c)}=\dfrac{\Delta^3\cdot s}{\Delta^2}=\Delta s=rs^2 \) (using \( (s-a)(s-b)(s-c)=\tfrac{\Delta^2}s \)).
2. \( r_1r_2+r_2r_3+r_3r_1=s^2 \), so \( r(r_1r_2+r_2r_3+r_3r_1)=rs^2 \).
3. Difference \( =rs^2-rs^2=0 \).

Teacher's Note:
1. Write each exradius as Delta divided by s minus a side.
2. Heron's formula gives (s-a)(s-b)(s-c) = Delta^2/s.