A Collection of Geometry Problems

by SomeonecoolLovesMaths, Dec 6, 2024, 9:54 AM

Here is a collection of geometry problems and solutions for new ideas:

Q1

In a triangle $ABC$ , the altitude $AD$ and the median $AE$ divide $\angle A$ into three equal parts. If $BC = 28$ , then the nearest integer to $AB + AC$ is?

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.92, xmax = 8.92, ymin = -5.4, ymax = 5.4; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((-3.26,3.82)--(-6.92,-1.34)--(6.76,-1.62)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw((-3.26,3.82)--(-6.92,-1.34), linewidth(2) + zzttqq); draw((-6.92,-1.34)--(6.76,-1.62), linewidth(2) + zzttqq); draw((6.76,-1.62)--(-3.26,3.82), linewidth(2) + zzttqq); draw((xmin, 48.857142857142854*xmin + 163.0942857142857)--(xmax, 48.857142857142854*xmax + 163.0942857142857), linewidth(2)); /* line */ draw((-3.26,3.82)--(-0.08,-1.48), linewidth(2)); /* dots and labels */ dot((-3.26,3.82),dotstyle); label("$A$", (-3.18,4.02), NE * labelscalefactor); dot((-6.92,-1.34),dotstyle); label("$B$", (-6.84,-1.14), NE * labelscalefactor); dot((6.76,-1.62),dotstyle); label("$C$", (6.84,-1.42), NE * labelscalefactor); label("$c$", (-5.36,1.44), NE * labelscalefactor,zzttqq); label("$a$", (-0.1,-1.78), NE * labelscalefactor,zzttqq); label("$b$", (1.9,1.4), NE * labelscalefactor,zzttqq); label("$f$", (-3.08,4.92), NE * labelscalefactor); dot((-3.36710245870117,-1.4127201251143038),linewidth(4pt) + dotstyle); label("$D$", (-3.28,-1.26), NE * labelscalefactor); dot((-0.08,-1.48),linewidth(4pt) + dotstyle); label("$E$", (0,-1.32), NE * labelscalefactor); label("$g$", (-1.96,1.02), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
$BE = EC = 14$ . $BD = DE = 7$ as $\triangle ABD \cong \triangle AED$ .
Using Internal Angle Bisector Theorem in $\triangle ADC$ .
$\frac{AD}{DE} = \frac{AC}{CE}$ , or, $2 AD = AC$ .
Using Pythagorean Theorem in $\triangle ADC$ ,
$21^2 + AD^2 = AC^2$ , or, $AD = \sqrt{147}$ .
Using Pythagorean Theorem in $\triangle ADB$ ,
$AB = \sqrt{x^2 + 7^2} = \sqrt{198}$ .
Thus $AB + AC = \sqrt{196} + 2 \sqrt{147} = 14(1+\sqrt{3}) \approx 38.3199585854$ .
Thus the closest integer is $\boxed{38}$ .

S1 by lbh_qys

Let $O$ be the circumcenter, then $\angle BAO = \angle CAD = \angle BAE$ , which indicates that $A$ , $O$ , and $E$ are colinear. Therefore, $O = E$ or $AB = AC$ . However, when $AB = AC$ , $AD$ and $AE$ coincide, and the angle is not trisected. Thus, $O = E$ , and triangle $ABC$ is a right-angled triangle. Consequently, the altitude divides the $90^\circ$ angle into $30^\circ$ and $60^\circ$ , which implies the triangle is a right-angled triangle with angles $30^\circ$ , $60^\circ$ , and $90^\circ$ . Therefore, $AB + AC = 14 + 14\sqrt{3} = 38.2\ldots$ , and the closest integer is $38$ .

The sides $a,b,c$ of a triangle satisfy the relations $c^2 = 2ab$ and $a^2 + c^2 = 3b^2$ . Then the measure of $\angle BAC$ , in degrees, is?

$a^2 + c^2 = a^2 + 2ab$ $a^2 + c^2 + b^2 = a^2 + 2ab + b^2 = (a+b)^2 = 4b^2 = (2b)^2$ $a+b = 2b$ $a = b$ $c^2 = 2a^2$ $c = \sqrt{2}a$ $\angle BAC = 45^{\circ}$

Let $PQR$ be a triangle in which $PQ=3$ . From the vertex $R$ , draw the altitude $RS$ to meet $PQ$ at $S$ . Assume that $RS = \sqrt{3}$ and $PS = QR$ . Then $PR$ equals?

Let $SQ = x$ . $QR = \sqrt{3+x^2} = PS = 3-x$ (Pythagorean Theorem). Thus $x=1$ and $RP = \sqrt{7}$ (Pythagorean Theorem).

Let $ABCD$ be a square of side length $1$ . Let $P,Q,R,S$ be points in the interiors of the sides $AD,BC,AB,CD$ , respectively, such that $PQ$ and $RS$ intersect at right angles. If $PQ = \frac{3 \sqrt{3}}{4}$ then find the length of $RS$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.92, xmax = 8.92, ymin = -5.4, ymax = 5.4; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((-3.58,2.26)--(-3.64,-3.22)--(1.84,-3.28)--(1.9,2.2)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw((-3.58,2.26)--(-3.64,-3.22), linewidth(2) + zzttqq); draw((-3.64,-3.22)--(1.84,-3.28), linewidth(2) + zzttqq); draw((1.84,-3.28)--(1.9,2.2), linewidth(2) + zzttqq); draw((1.9,2.2)--(-3.58,2.26), linewidth(2) + zzttqq); draw((-3.5946697742558436,0.9201606179663049)--(1.8594891123393478,-1.499994406339482), linewidth(2)); draw((xmin, 2.253640296517657*xmin + 0.49841400936201097)--(xmax, 2.253640296517657*xmax + 0.49841400936201097), linewidth(2)); /* line */ draw((0.760574548844643,2.2124754611440354)--(-1.6595804754611438,-3.2416834254511557), linewidth(2)); draw((xmin, -0.010948905109489145*xmin-1.47963503649635)--(xmax, -0.010948905109489145*xmax-1.47963503649635), linewidth(2)); /* line */ draw((xmin, 91.33333333333121*xmin + 148.3333333333298)--(xmax, 91.33333333333121*xmax + 148.3333333333298), linewidth(2)); /* line */ /* dots and labels */ dot((-3.58,2.26),dotstyle); label("$A$", (-3.5,2.46), NE * labelscalefactor); dot((-3.64,-3.22),dotstyle); label("$B$", (-3.56,-3.02), NE * labelscalefactor); dot((1.84,-3.28),dotstyle); label("$C$", (1.92,-3.08), NE * labelscalefactor); dot((1.9,2.2),dotstyle); label("$D$", (1.98,2.4), NE * labelscalefactor); dot((-3.5946697742558436,0.9201606179663049),dotstyle); label("$E$", (-3.52,1.12), NE * labelscalefactor); dot((1.8594891123393478,-1.499994406339482),dotstyle); label("$F$", (1.94,-1.3), NE * labelscalefactor); label("$j$", (-1,-0.58), NE * labelscalefactor); dot((-1.6595804754611438,-3.2416834254511557),dotstyle); label("$G$", (-1.58,-3.04), NE * labelscalefactor); label("$k$", (1.74,4.92), NE * labelscalefactor); dot((0.760574548844643,2.2124754611440354),linewidth(4pt) + dotstyle); label("$H$", (0.84,2.38), NE * labelscalefactor); label("$l$", (-0.76,-0.38), NE * labelscalefactor); label("$m$", (-8.76,-1.22), NE * labelscalefactor); label("$n$", (-1.42,4.92), NE * labelscalefactor); dot((-1.5995804754611425,2.238316574548844),linewidth(4pt) + dotstyle); label("$I$", (-1.52,2.4), NE * labelscalefactor); dot((-3.620510887660651,-1.4399944063394812),linewidth(4pt) + dotstyle); label("$J$", (-3.54,-1.28), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

We rename the points as per the diagram.
Let $FJ || AD$ and $GI || AB$ .
Then, $\triangle GIH \cong \triangle FJE$ as $\angle EJF = \angle HIG = 90^{\circ}$ , and $FJ = GI$ and $\angle EFJ = \angle HGI$ .
Thus the required length is $\frac{3 \sqrt{3}}{4}$ .

S4 by Mathzeus1024

Let the points $A(0,L); B(L,L); C(L,0); D(0,0); P(0,p); Q(L,q); R(r, L); S(s,0)$ be located in the $xy-$ plane. If $PQ \perp RS$ , then:

$m_{PQ} \cdot m_{RS} =-1 \Rightarrow \frac{p-q}{0-L} = -\frac{r-s}{L-0} \Rightarrow p-q=r-s$ (i).

If $|PQ| = \frac{3\sqrt{3}}{4}$ , then $\sqrt{(0-L)^2+(p-q)^2} = \frac{3\sqrt{3}}{4} \Rightarrow p-q = \sqrt{\frac{27}{16}-L^2} =r-s$ (ii).

Finally, $|RS| = \sqrt{(r-s)^2+(L-0)^2} =\sqrt{\left(\sqrt{\frac{27}{16}-L^2}\right)^2 + L^2} = \textcolor{red}{\frac{3\sqrt{3}}{4}}$ .

Let $P$ be a point inside a triangle $ABC$ with $\angle ABC = 90^{\circ}$ . Let $P_1$ and $P_2$ be the images of $P$ under reflection in $AB$ and $BC$ respectively. The distance between the circumcenters of triangles $ABC$ and $P_1P P_2$ is?
$\textbf{(A)} \frac{AB}{2} ~\textbf{(B)} \frac{AP+BP+CP}{3} ~\textbf{(C)} \frac{AC}{2} ~\textbf{(D)} \frac{AB+BC+CA}{2}$

Let $B = (0,0)$ and $A = (0,a)$ and $C = (c,0)$ . Assume $P = (m,n)$ then $P_1 = (-m,n)$ and $P_2 = (m,-n)$ . Observe that the circumcenter of $\triangle ABC$ is just $\left( \frac{c}{2} , \frac{a}{2}\right)$ . Observe that the circumcenter of $\triangle P_1PP_2$ is just $(0,0) = B$ . Thus the distance is just the circumradius or $\frac{AC}{2} = \textbf(C)$ .

Three circles of radii $1,2$ and $3$ units respectively touch each other externally in the place. The circumradius of the triangle formed by joining the centers of the circles is?

Observe the triangle formed is a $3,4,5$ right angled triangle thus the circumradius is $\frac{5}{2} = 2.5$ .

Given $\triangle ABC$ with $\angle A = 15^{\circ}$ , let $M$ be midpoint of $BC$ and let $E$ and $F$ be points on ray
$BA$ and $CA$ respectively such that $BE = BM = CF$ . Let $R_1$ be the radius of $(MEF)$ and $R_{2}$ be
radius of $(AEF)$ . If $\frac{R_1^2}{R_2^2}=a-\sqrt{b+\sqrt{c}}$ where $a,b,c$ are integers. Find $a^{b^{c}}$

Let $\angle BEM = \alpha = \angle BME$ and $\angle CFB = \theta = \angle CBF$ .
Let $\angle AEF = \gamma$ and $\angle AFE = 165^{\circ} - \gamma$ .
Thus, $\angle FEB = 180^{\circ} - \alpha - \gamma$ and $\angle EFB = 15 + \alpha - \theta$ .
Thus, $\angle EMF = \alpha + \theta - 15^{\circ}$ .
Thus, $180^{\circ} = 2( \alpha + \theta) - 15^{circ}$ or $\angle EFM = 82.5^{\circ}$ .
Using LoS on $EF$ ,
In $(AEF)$ using LoS, $\frac{EF}{\sin (15^{\circ})} = 2R_2$ .
In $(MEF)$ using LoS, $\frac{EF}{\sin (82.5^{\circ})} = 2R_1$ .
Thus, $\frac{R_1 ^2}{R_2 ^2} = \frac{\sin ^2 (15^{\circ})}{ \cos ^2 (7.5 ^{\circ})} = 4 \sin ^2 (7.5 ^{\circ}) = 2 - \frac{ \sqrt{6} - \sqrt{2}}{2} = 2 - \sqrt{2 + \sqrt{3}}$ .
Thus, the final answer is $2^{2^{3}} = \boxed{256}$ .

Let $ABC$ be a triangle in which $AB = AC$ and let $I$ be its in-centre. Suppose $BC = AB + AI$ . Find $\angle{BAC}$

Construct $AD$ from $AB$ exterior to $\bigtriangleup ABC$ such that $AI = AD$ .
Note that $\triangle BDC$ is isosceles.
Let $\angle ABC = \angle ACB = \theta$ .
Thus $\angle AIC = 90^{\circ} - \theta$ and $\angle CAD = 2 \theta$ . This means that $\angle AID = \angle ADI = 45^{\circ} - \frac{\theta}{2}$ (as $\triangle AID$ is isosceles.
As $\triangle BCD$ is isosceles, $\angle BCD = \angle BDC$ which means that $\angle IDC = 45^{\circ}$ and $\angle ACD = 90^{\circ} - \frac{3 \theta}{2}$ .
As $\angle ACI = \frac{\theta}{2}$ .
Observe $\angle IAD + \angle ICD = 90^{\circ} + \theta + 90^{\circ} - \theta - 180^{\circ}$ . Thus $AICD$ is a cyclic quadrilateral.
Thus $\angle AID = \angle ACD$ which means that $45^{\circ} - \frac{\theta}{2} = 90^{\circ} - \frac{3 \theta}{2}$ . Thus $\theta = 45^{\circ}$ .
$\angle BAC = 180^{\circ} - 2 \theta = 90^{\circ}$ .

In $\triangle ABC$ , $\angle A=30^{\circ}$ , $BC=13$ . Given $2$ circles $\gamma_1, \gamma_2$ with radius $r_1,r_2$ contain $A$ and touch $BC$ at $B$ and $C$ respectively. Find $r_1r_2$ .

Let $D,E$ be the midpoints of $AB,AC$ respectively. Let $O_1,O_2$ be the centres of $\gamma_1, \gamma_2$ respectively.
Thus, $\angle O_1 BD = 90^{\circ} - \angle B \Longrightarrow \cos (\angle O_1BD) = \frac{BD}{O_1 B } = \frac{AB}{2 r_1}$ $\cos (\angle O_1BD) = \sin ( \angle B O_1 D) = \sin (\angle B)$ Similarly, $\sin (\angle C) = \frac{AC}{2 r_1}$ Using LoS, $\frac{ \sin 30^{\circ}}{BC} = \frac{\sin (\angle B)}{AC} = \frac{\sin (\angle C)}{AB}$ $\frac{1}{2 BC} = \frac{\sin (\angle B)}{AC} = \frac{AB}{2 r_1 AC}$ or $r_1 = \frac{AB}{AC} \times BC$ .
Similarly, $r_2 = \frac{AC}{AB} \times BC$ .
$\therefore r_1 r_2 = BC^2 = \boxed{169}$

This post has been edited 12 times. Last edited by SomeonecoolLovesMaths, Jan 8, 2025, 4:47 PM

My First Blog

A Collection of Geometry Problems

by SomeonecoolLovesMaths, Dec 6, 2024, 9:54 AM

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